3.1 Complex number calculations

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{{Vald flik|[[3.1 Räkning med komplexa tal|Teori]]}}
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{{Selected tab|[[3.1 Complex number calculations|Theory]]}}
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{{Ej vald flik|[[3.1 Övningar|Övningar]]}}
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{{Not selected tab|[[3.1 Exercises|Exercises]]}}
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{{Info|
{{Info|
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'''Innehåll:'''
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'''Contents:'''
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* Real- och imaginärdel
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* Real and imaginary part
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* Addition och subtraktion av komplexa tal
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* Addition and subtraction of complex numbers
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* Komplexkonjugat
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* Complex conjugate
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* Multiplikation och division av komplexa tal
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* Multiplication and division of complex numbers
}}
}}
{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes:'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned to:
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* Beräkna uttryck som innehåller komplexa tal och är uppbyggda av de fyra räknesätten.
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* Simplify expressions that are constructed using complex numbers and the four arithmetic operations.
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* Lösa komplexa förstagradsekvationer och förenkla svaret.
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* Solve first order complex number equations and simplify the answer.
}}
}}
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== Inledning ==
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== Introduction ==
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De reella talen utgör en fullständig mängd av tal i den meningen att de fyller tallinjen, dvs. det finns inga "hål" i den reella tallinjen. Trots detta räcker de reella talen inte till som lösningar till alla algebraiska ekvationer, dvs. det finns ekvationer av typen
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The real numbers represent a complete set of numbers in the sense that they "fill" the real-number axis. Despite this, the set of real numbers does not contain the solutions of ''all'' possible algebraic equations. In other words, there are equations of the type
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{{Fristående formel||<math>a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0</math>}}
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{{Displayed math||<math>a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0</math>}}
-
som inte har någon lösning bland de reella talen. Exempelvis har ekvationen <math>x^2+1=0</math> ingen reell lösning, eftersom inget reellt tal uppfyller att <math>x^2=-1</math>. Om vi däremot kan tänka oss <math>\sqrt{-1}</math> som det tal som uppfyller ekvationen <math>x^2=-1</math> och tillåter oss att räkna med <math>\sqrt{-1}</math> som vilket tal som helst, så visar det sig att alla algebraiska ekvationer har lösningar.
+
that do not have a solution among the real numbers. For example, the equation <math>x^2+1=0</math> has no real solution, since no real number satisfies <math>x^2=-1</math>. However, if we can imagine <math>\sqrt{-1}</math> as the number that satisfies the equation <math>x^2=-1</math> and manipulate <math>\sqrt{-1}</math> just like any other number, it turns out that every algebraic equation does have solutions.
-
 
+
The number <math>\sqrt{-1}</math>, however, is a fairly strange object. We cannot go out into the world and measure <math>\sqrt{-1}</math> anywhere, or find something that is numerically <math>\sqrt{-1}</math>. Nonetheless, this number turns out to be very useful in many applications of mathematics.
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Talet <math>\sqrt{-1}</math> är alltså inget reellt tal; vi kan inte gå ut i naturen och uppmäta <math>\sqrt{-1}</math> någonstans, eller hitta något som är <math>\sqrt{-1}</math> till antalet, men vi kan ändå ha nytta av talet i högst reella sammanhang.
+
<div class="exempel">
<div class="exempel">
-
'''Exempel 1'''
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''' Example 1'''
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Om vi skulle vilja ta reda på summan av rötterna (lösningarna) till ekvationen <math>x^2-2x+2=0</math> så får vi först lösningarna <math>x_1=1+\sqrt{-1}</math> och <math>x_2=1-\sqrt{-1}</math>. Dessa rötter innehåller det icke-reella talet <math>\sqrt{-1}</math>. Om vi för en stund tillåter oss att räkna med <math>\sqrt{-1}</math> så ser vi att summan av <math>x_1</math> och <math>x_2</math> blir <math>1+\sqrt{-1} + 1-\sqrt{-1} =2</math>, alltså ett högst reellt tal.
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If we would like to find out the sum of the roots (solutions) of the equation <math>x^2-2x+2=0</math> we can first obtain the roots <math>x_1=1+\sqrt{-1}</math> and <math>x_2=1-\sqrt{-1}</math>. These roots contain <math>\sqrt{-1}</math>. If we allow ourselves to do calculations containing <math>\sqrt{-1}</math>, we see that the sum of <math>x_1</math> and <math>x_2</math> turns out to be <math>1+\sqrt{-1} + 1-\sqrt{-1} =2</math>, which is an ordinary, familiar "real" number.
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För att lösa vårt problem var vi här tvungna att tillfälligtvis använda tal som inte är reella för att komma fram till den reella lösningen.
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Even though the answer to the problem was a "real" number, we found the "imaginary" number <math>\sqrt{-1}</math> useful in solving it.
</div>
</div>
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== Definition av komplexa tal ==
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== Definition of complex numbers==
 +
 
 +
The terms "real" (for the ordinary, familiar positive and negative numbers, together with zero) and "imaginary" (for numbers like <math>\sqrt{-1}</math>) are actually pretty flawed in some ways (''all'' numbers are abstract human inventions, it could be argued). Nonetheless, this terminology, which reflects the unease and scepticism with which the mathematical community once viewed numbers like <math>\sqrt{-1}</math>, has stuck.
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Man inför den ''imaginära enheten'' <math>i=\sqrt{-1}</math> och definierar ett ''komplext tal'' som ett objekt som kan skrivas på formen
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The number <math>\sqrt{-1}</math> cannot be given an approximate decimal value, the way <math>\sqrt{2}</math> can; we have no choice but to write it as an unevaluated root (a ''surd''). For simplicity and economy, we usually represent this number by the symbol <math>i</math> (sometimes <math>j</math>). It is sometimes called the ''imaginary unit'', and any number of the form <math>b\,i</math>, where <math>b</math> is real, is known as an ''imaginary number''. We go on to define a ''complex number'' as an object that can be written in the form
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>z=a+bi\,\mbox{,}</math>}}
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{{Displayed math||<math>z=a+bi\,\mbox{,}</math>}}
</div>
</div>
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där <math>a</math> och <math>b</math> är reella tal, och <math>i</math> uppfyller <math>i^2=-1</math>.
+
where <math>a</math> and <math>b</math> are real numbers, and <math>i</math> satisfies <math>i^2=-1</math>.
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Om <math>a = 0</math> så kallas talet "rent imaginärt". Om <math>b = 0</math> så är talet reellt. Vi ser här att de reella talen utgör en delmängd av de komplexa talen. Mängden av de komplexa talen betecknas med '''C'''.
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If <math>a = 0</math> then the number is "purely imaginary". If <math>b = 0</math> the number is real. We can see that the real numbers are a subset of the complex numbers. The set of complex numbers is designated by '''C'''.
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För ett godtyckligt komplext tal använder man ofta beteckningen <math>z</math>. Om <math>z=a+bi</math>, där <math>a</math> och <math>b</math> är reella, så kallas <math>a</math> för realdelen och <math>b</math> för imaginärdelen av <math>z</math>. Man använder följande skrivsätt:<br\>
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For an arbitrary complex number one often uses the symbol <math>z</math>. If <math>z=a+bi</math>, where <math>a</math> and <math>b</math> are real, then <math>a</math> is the real part and <math>b</math> the imaginary part of <math>z</math>. One uses the following notation: <br\>
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>\begin{align*}a &= \mathop{\rm Re} z\,\mbox{,}\\ b&=\mathop{\rm Im} z\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*}a &= \mathop{\rm Re} z\,\mbox{,}\\ b&=\mathop{\rm Im} z\,\mbox{.}\end{align*}</math>}}
</div>
</div>
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När man räknar med komplexa tal gör man i princip som med de reella talen, men håller reda på att <math>i^2=-1</math>.
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When one calculates with complex numbers one treats them just like real numbers, but keeps track of the fact that <math>i^2=-1</math>.
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== Addition och subtraktion ==
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==Addition and subtraction ==
-
Vid addition och subtraktion av komplexa tal lägger man ihop (drar ifrån) realdel och imaginärdel var för sig. Om <math>z=a+bi</math> och <math>w=c+di</math> är två komplexa tal gäller alltså att
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To add or subtract complex numbers one adds or subtracts the real and imaginary parts separately.
 +
If <math>z=a+bi</math> and <math>w=c+di</math> are two complex numbers then,
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>\begin{align*} z+w &= a+bi+c+di = a+c+(b+d)i\,\mbox{,}\\ z-w &= a+bi-(c+di) = a-c+(b-d)i\,\mbox{.}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*} z+w &= a+bi+c+di = a+c+(b+d)i\,\mbox{,}\\ z-w &= a+bi-(c+di) = a-c+(b-d)i\,\mbox{.}\end{align*}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
<ol type="a">
<ol type="a">
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= \tfrac{1}{3}-i</math></li>
= \tfrac{1}{3}-i</math></li>
<li> <math>\frac{3+2i}{5}-\frac{3-i}{2} = \frac{6+4i}{10}-\frac{15-5i}{10}
<li> <math>\frac{3+2i}{5}-\frac{3-i}{2} = \frac{6+4i}{10}-\frac{15-5i}{10}
-
= \frac{-9+9i}{10} = -0{,}9 + 0{,}9i</math></li>
+
= \frac{-9+9i}{10} = -0\textrm{.}9 + 0\textrm{.}9i</math></li>
</ol>
</ol>
</div>
</div>
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== Multiplikation ==
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== Multiplication ==
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Komplexa tal multipliceras som vanliga reella tal eller algebraiska uttryck, med tillägget att <math>i^2=-1</math>. Generellt gäller för två komplexa tal <math>z=a+bi</math> och <math>w=c+di</math> att
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Complex numbers are multiplied in the same way as ordinary real numbers or algebraic expressions, with the extra condition that <math>i^2=-1</math>. Generally one has for two complex numbers <math>z=a+bi</math> and <math>w=c+di</math> that
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<div class="regel">{{Fristående formel||<math>z\cdot w=(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i\,\mbox{.}</math>}}
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<div class="regel">{{Displayed math||<math>z\, w=(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i\,\mbox{.}</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 3'''
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''' Example 3'''
<ol type="a">
<ol type="a">
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<li><math>(1+i)(2+3i)=2+3i+2i+3i^2=-1+5i</math></li>
<li><math>(1+i)(2+3i)=2+3i+2i+3i^2=-1+5i</math></li>
<li><math>(3+2i)(3-2i)=3^2-(2i)^2=9-4i^2=13</math></li>
<li><math>(3+2i)(3-2i)=3^2-(2i)^2=9-4i^2=13</math></li>
-
<li><math>(3+i)^2=3^2+2\cdot3i+i^2=8+6i</math></li>
+
<li><math>(3+i)^2=3^2+2\times3i+i^2=8+6i</math></li>
<li><math>i^{12}=(i^2)^6=(-1)^6=1</math></li>
<li><math>i^{12}=(i^2)^6=(-1)^6=1</math></li>
-
<li><math>i^{23}=i^{22}\cdot i=(i^2)^{11}\cdot i=(-1)^{11}i=-i</math></li>
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<li><math>i^{23}=i^{22}\times i=(i^2)^{11}\times i=(-1)^{11}i=-i</math></li>
</ol>
</ol>
</div>
</div>
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== Komplexkonjugat ==
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== Complex conjugate ==
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Om <math>z=a+bi</math> så kallas <math>\overline{z} = a-bi</math> det ''komplexa konjugatet'' till <math>z</math> (omvänt gäller också att <math>z</math> är konjugatet till <math>\overline{z}</math>). Man får då sambanden
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If <math>z=a+bi</math> then <math>\overline{z} = a-bi</math> is called the ''complex conjugate'' of <math>z</math> (the opposite is also true, that <math>z</math> is conjugate to <math>\overline{z}</math>). One obtains the relationships
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>\begin{align*} z+\overline{z} &= a + bi + a - bi = 2a = 2\, \mathop{\rm Re} z\,\mbox{,}\\ z-\overline{z} &= a + bi - (a - bi) = 2bi = 2i\, \mathop{\rm Im} z\,\mbox{,}\end{align*}</math>}}
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{{Displayed math||<math>\begin{align*} z+\overline{z} &= a + bi + a - bi = 2a = 2\, \mathop{\rm Re} z\,\mbox{,}\\ z-\overline{z} &= a + bi - (a - bi) = 2bi = 2i\, \mathop{\rm Im} z\,\mbox{,}\end{align*}</math>}}
</div>
</div>
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men kanske framför allt, pga. konjugatregeln, att
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but most importantly, using the difference of two squares rule, one obtains
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>z\cdot \bar z = (a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2=a^2+b^2\,\mbox{,}</math>}}
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{{Displayed math||<math>z\, \bar z = (a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2=a^2+b^2\,\mbox{,}</math>}}
</div>
</div>
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dvs. att produkten av ett komplext tal och dess konjugat är alltid reell.
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i.e. that the product of a complex number and its conjugate is always real.
 +
 
<div class="exempel">
<div class="exempel">
-
'''Exempel 4'''
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''' Example 4'''
<ol type="a">
<ol type="a">
-
<li><math>z=5+i\qquad</math> då är <math>\quad\overline{z}=5-i\,</math>.</li>
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<li><math>z=5+i\qquad</math> then <math>\quad\overline{z}=5-i\,</math>.</li>
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<li><math>z=-3-2i\qquad</math> då är <math>\quad\overline{z} =-3+2i\,</math>.</li>
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<li><math>z=-3-2i\qquad</math> then <math>\quad\overline{z} =-3+2i\,</math>.</li>
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<li><math>z=17\qquad</math> då är <math>\quad\overline{z} =17\,</math>.</li>
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<li><math>z=17\qquad</math> then <math>\quad\overline{z} =17\,</math>.</li>
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<li><math>z=i\qquad</math> då är <math>\quad\overline{z} =-i\,</math>.</li>
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<li><math>z=i\qquad</math> then <math>\quad\overline{z} =-i\,</math>.</li>
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<li><math>z=-5i\qquad</math> då är <math>\quad\overline{z} =5i\,</math>.</li>
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<li><math>z=-5i\qquad</math> then <math>\quad\overline{z} =5i\,</math>.</li>
</ol>
</ol>
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 5'''
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''' Example 5'''
<ol type="a">
<ol type="a">
-
<li> Om <math>z=4+3i</math> då gäller att
+
<li> If <math>z=4+3i</math> one has
*<math>z+\overline{z} = 4 + 3i + 4 -3i = 8</math>
*<math>z+\overline{z} = 4 + 3i + 4 -3i = 8</math>
*<math>z-\overline{z} = 6i</math>
*<math>z-\overline{z} = 6i</math>
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*<math>z \cdot \overline{z} = 4^2-(3i)^2=16+9=25</math>
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*<math>z \, \overline{z} = 4^2-(3i)^2=16+9=25</math>
</li>
</li>
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<li> För <math>z</math> gäller att <math>\mathop{\rm Re} z=-2</math>
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<li> If for <math>z</math> one has <math>\mathop{\rm Re} z=-2</math>
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och <math>\mathop{\rm Im} z=1</math>, och får vi att
+
and <math>\mathop{\rm Im} z=1</math>, one gets
*<math>z+\overline{z} = 2\,\mathop{\rm Re} z = -4</math>
*<math>z+\overline{z} = 2\,\mathop{\rm Re} z = -4</math>
*<math>z-\overline{z} = 2i\,\mathop{\rm Im} z = 2i</math>
*<math>z-\overline{z} = 2i\,\mathop{\rm Im} z = 2i</math>
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*<math>z\cdot \overline{z} = (-2)^2+1^2=5</math>
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*<math>z\, \overline{z} = (-2)^2+1^2=5</math>
</li>
</li>
</ol>
</ol>
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== Division ==
== Division ==
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När man dividerar två komplexa tal med varandra förlänger man med nämnarens konjugat och utnyttjar härigenom att nämnaren då blir ett reellt tal. Därefter kan såväl realdel som imaginärdel i täljaren divideras med detta tal. Generellt, om <math>z=a+bi</math> och <math>w=c+di</math>:
+
For the division of two complex numbers one multiplies the numerator and denominator by the complex conjugate of the denominator, thus getting a denominator which is a real number. Then, both the real and imaginary parts of the (new) numerator are divided by this number (the new denominator). In general, if <math>z=a+bi</math> and <math>w=c+di</math>:
<div class="regel">
<div class="regel">
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{{Fristående formel||<math>\frac{z}{w} = \frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}\,i</math>}}
+
{{Displayed math||<math>\frac{z}{w} = \frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}\,i</math>}}
</div>
</div>
<div class="exempel">
<div class="exempel">
-
'''Exempel 6'''
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''' Example 6'''
<ol type="a">
<ol type="a">
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<div class="exempel">
<div class="exempel">
-
'''Exempel 7'''
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''' Example 7'''
<ol type="a">
<ol type="a">
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= \frac{\dfrac{-1-i}{1-i}}{\dfrac{-2+5i}{2+i}}</math><br/>
= \frac{\dfrac{-1-i}{1-i}}{\dfrac{-2+5i}{2+i}}</math><br/>
<math>\quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{}
<math>\quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{}
-
= \frac{-1-i}{1-i}\cdot \frac{2+i}{-2+5i}
+
= \frac{-1-i}{1-i}\times \frac{2+i}{-2+5i}
= \frac{(-1-i)(2+i)}{(1-i)(-2+5i)}
= \frac{(-1-i)(2+i)}{(1-i)(-2+5i)}
= \frac{-2-i-2i-i^2}{-2+5i+2i-5i^2}\vphantom{\Biggl(}</math><br/>
= \frac{-2-i-2i-i^2}{-2+5i+2i-5i^2}\vphantom{\Biggl(}</math><br/>
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<div class="exempel">
<div class="exempel">
-
'''Exempel 8'''
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''' Example 8'''
-
Bestäm det reella talet <math>a</math> så att uttrycket <math>\ \frac{2-3i}{2+ai}\ </math> blir reellt.
+
Determine the real number <math>a</math> such that the expression <math>\ \frac{2-3i}{2+ai}\ </math> becomes real.
<br\>
<br\>
<br\>
<br\>
-
Förläng med nämnarens konjugat så att uttrycket kan skrivas med separata real- och imaginärdelar
+
Multiply the numerator and denominator by the complex conjugate of the denominator so that the expression can be written with separate real and imaginary parts.
-
{{Fristående formel||<math>\frac{(2-3i)(2-ai)}{(2+ai)(2-ai)} = \frac{4-2ai-6i+3ai^2}{4-a^2i^2} = \frac{4-3a-(2a+6)i}{4+a^2}</math>}}
+
{{Displayed math||<math>\frac{(2-3i)(2-ai)}{(2+ai)(2-ai)} = \frac{4-2ai-6i+3ai^2}{4-a^2i^2} = \frac{4-3a-(2a+6)i}{4+a^2}</math>}}
-
Om uttrycket ska bli reellt så måste imaginärdelen vara 0, dvs.
+
If the expression is to be real , the imaginary part must be 0, ie.
-
{{Fristående formel||<math>2a+6=0\quad\Leftrightarrow\quad a = -3\,\mbox{.}</math>}}
+
{{Displayed math||<math>2a+6=0\quad\Leftrightarrow\quad a = -3\,\mbox{.}</math>}}
</div>
</div>
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== Ekvationer ==
+
== Equations ==
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För att två komplexa tal <math>z=a+bi</math> och <math>w=c+di</math> ska vara lika, krävs att både real- och imaginärdel är lika, dvs. att <math>a=c</math> och <math>b=d</math>. När man söker ett okänt komplext tal <math>z</math> i en ekvation kan man antingen försöka lösa ut talet <math>z</math> på vanligt vis, eller sätta in <math>z=a+bi</math> i ekvationen och därefter jämföra real- och imaginärdelar i ekvationens båda led med varandra.
+
If the two complex numbers <math>z=a+bi</math> and <math>w=c+di</math> are equal then it is not hard to show that both the real and imaginary parts must be equal (in other words, <math>a=c</math> and <math>b=d</math>). When you are looking for an unknown complex number <math>z</math> in an equation, you can either try to solve for the number <math>z</math> in the usual way, or insert <math>z=a+bi</math> in the equation and then compare the real and imaginary parts of the two sides of the equation with each other.
<div class="exempel">
<div class="exempel">
-
'''Exempel 9'''
+
''' Example 9'''
<ol type="a">
<ol type="a">
-
<li> Lös ekvationen <math>3z+1-i=z-3+7i</math>.
+
<li> Solve the equation <math>3z+1-i=z-3+7i</math>.
<br/>
<br/>
<br/>
<br/>
-
Samla <math>z</math> i vänsterledet genom att subtrahera båda led med <math>z</math>
+
Collect all <math>z</math> on the left-hand side by subtracting <math>z</math> from both sides
-
{{Fristående formel||<math>2z+1-i = -3+7i</math>}}
+
{{Displayed math||<math>2z+1-i = -3+7i</math>}} and now subtract <math>1-i</math>
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och subtrahera sedan med <math>1-i</math>
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{{Displayed math||<math>2z = -4+8i\,\mbox{.}</math>}}
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{{Fristående formel||<math>2z = -4+8i\,\mbox{.}</math>}}
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This gives that <math>\ z=\frac{-4+8i}{2}=-2+4i\,\mbox{.}</math></li>
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Detta ger att <math>\ z=\frac{-4+8i}{2}=-2+4i\,\mbox{.}</math></li>
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<li> Lös ekvationen <math>z(-1-i)=6-2i</math>.
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<li> Solve the equation <math>z(-1-i)=6-2i</math>.
<br/>
<br/>
<br/>
<br/>
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Dela båda led med <math>-1-i</math> för att få fram <math>z</math>
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Divide both sides by <math>-1-i</math> in order to obtain <math>z</math>
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{{Fristående formel||<math>z = \frac{6-2i}{-1-i} = \frac{(6-2i)(-1+i)}{(-1-i)(-1+i)} = \frac{-6+6i+2i-2i^2}{(-1)^2 -i^2} = \frac{-4+8i}{2} = -2+4i\,\mbox{.}</math>}}</li>
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{{Displayed math||<math>z = \frac{6-2i}{-1-i} = \frac{(6-2i)(-1+i)}{(-1-i)(-1+i)} = \frac{-6+6i+2i-2i^2}{(-1)^2 -i^2} = \frac{-4+8i}{2} = -2+4i\,\mbox{.}</math>}}</li>
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<li> Lös ekvationen <math>3iz-2i=1-z</math>.
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<li> Solve the equation <math>3iz-2i=1-z</math>.
<br/>
<br/>
<br/>
<br/>
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Adderar vi <math>z</math> och <math>2i</math> till båda led fås
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Adding <math>z</math> and <math>2i</math> to both sides gives
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{{Fristående formel||<math>3iz+z=1+2i\quad \Leftrightarrow \quad z(3i+1)=1+2i\,\mbox{.}</math>}}
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{{Displayed math||<math>3iz+z=1+2i\quad \Leftrightarrow \quad z(3i+1)=1+2i\,\mbox{.}</math>}}
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Detta ger att
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This gives
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{{Fristående formel||<math>z = \frac{1+2i}{1+3i} = \frac{(1+2i)(1-3i)}{(1+3i)(1-3i)} = \frac{1-3i+2i-6i^2}{1-9i^2} = \frac{7-i}{10}\,\mbox{.}</math>}}</li>
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{{Displayed math||<math>z = \frac{1+2i}{1+3i} = \frac{(1+2i)(1-3i)}{(1+3i)(1-3i)} = \frac{1-3i+2i-6i^2}{1-9i^2} = \frac{7-i}{10}\,\mbox{.}</math>}}</li>
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<li> Lös ekvationen <math>2z+1-i=\bar z +3 + 2i</math>.
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<li> Solve the equation <math>2z+1-i=\bar z +3 + 2i</math>.
<br/>
<br/>
<br/>
<br/>
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I ekvationen förekommer <math>z</math> också som <math>\overline{z}</math> och därför skriver vi <math>z</math> som <math>z=a+ib</math> och löser ekvationen för <math>a</math> och <math>b</math> genom att sätta real- och imaginärdel av båda led lika
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The equation contains both <math>z</math> as well as <math>\overline{z}</math> and therefore we assume <math>z</math> to be <math>z=a+ib</math> and solve the equation for <math>a</math> and <math>b</math> by equating the real and imaginary parts of both sides {{Displayed math||<math>2(a+bi)+1-i=(a-bi)+3+2i</math>}}
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{{Fristående formel||<math>2(a+bi)+1-i=(a-bi)+3+2i</math>}}
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i.e.
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dvs.
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{{Displayed math||<math>(2a+1)+(2b-1)i=(a+3)+(2-b)i\,\mbox{,}</math>}}
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{{Fristående formel||<math>(2a+1)+(2b-1)i=(a+3)+(2-b)i\,\mbox{,}</math>}}
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which gives
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vilket ger att
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{{Displayed math||<math>\left\{\begin{align*} 2a+1 &= a+3\\ 2b-1 &= 2-b\end{align*}\right.\quad\Leftrightarrow\quad\left\{\begin{align*} a &= 2\\ b &= 1\end{align*}\right.\,\mbox{.}</math>}}
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{{Fristående formel||<math>\left\{\begin{align*} 2a+1 &= a+3\\ 2b-1 &= 2-b\end{align*}\right.\quad\Leftrightarrow\quad\left\{\begin{align*} a &= 2\\ b &= 1\end{align*}\right.\,\mbox{.}</math>}}
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The answer is therefore, <math>z=2+i</math>.</li>
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Svaret är alltså <math>z=2+i</math>.</li>
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</ol>
</ol>
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<div class="inforuta">
<div class="inforuta">
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'''Råd för inläsning'''
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'''Study advice '''
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'''Tänk på att:'''
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Räkning med komplexa tal fungerar på samma sätt som med vanliga tal förutom att <math>i^2=-1</math>.
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'''Keep in mind that:'''
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Kvoter av komplexa tal räknas ut genom att förlänga bråket med nämnarens komplexkonjugat.
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Calculations with complex numbers are done in the same way as with ordinary numbers with the additional information that <math>i^2=-1</math>.
 +
Quotients of complex numbers are simplified by multiplying the numerator and denominator by the complex conjugate of the denominator.
</div>
</div>

Current revision

       Theory          Exercises      

Contents:

  • Real and imaginary part
  • Addition and subtraction of complex numbers
  • Complex conjugate
  • Multiplication and division of complex numbers

Learning outcomes:

After this section, you will have learned to:

  • Simplify expressions that are constructed using complex numbers and the four arithmetic operations.
  • Solve first order complex number equations and simplify the answer.


Introduction

The real numbers represent a complete set of numbers in the sense that they "fill" the real-number axis. Despite this, the set of real numbers does not contain the solutions of all possible algebraic equations. In other words, there are equations of the type

\displaystyle a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0

that do not have a solution among the real numbers. For example, the equation \displaystyle x^2+1=0 has no real solution, since no real number satisfies \displaystyle x^2=-1. However, if we can imagine \displaystyle \sqrt{-1} as the number that satisfies the equation \displaystyle x^2=-1 and manipulate \displaystyle \sqrt{-1} just like any other number, it turns out that every algebraic equation does have solutions. The number \displaystyle \sqrt{-1}, however, is a fairly strange object. We cannot go out into the world and measure \displaystyle \sqrt{-1} anywhere, or find something that is numerically \displaystyle \sqrt{-1}. Nonetheless, this number turns out to be very useful in many applications of mathematics.


Example 1

If we would like to find out the sum of the roots (solutions) of the equation \displaystyle x^2-2x+2=0 we can first obtain the roots \displaystyle x_1=1+\sqrt{-1} and \displaystyle x_2=1-\sqrt{-1}. These roots contain \displaystyle \sqrt{-1}. If we allow ourselves to do calculations containing \displaystyle \sqrt{-1}, we see that the sum of \displaystyle x_1 and \displaystyle x_2 turns out to be \displaystyle 1+\sqrt{-1} + 1-\sqrt{-1} =2, which is an ordinary, familiar "real" number.

Even though the answer to the problem was a "real" number, we found the "imaginary" number \displaystyle \sqrt{-1} useful in solving it.


Definition of complex numbers

The terms "real" (for the ordinary, familiar positive and negative numbers, together with zero) and "imaginary" (for numbers like \displaystyle \sqrt{-1}) are actually pretty flawed in some ways (all numbers are abstract human inventions, it could be argued). Nonetheless, this terminology, which reflects the unease and scepticism with which the mathematical community once viewed numbers like \displaystyle \sqrt{-1}, has stuck.

The number \displaystyle \sqrt{-1} cannot be given an approximate decimal value, the way \displaystyle \sqrt{2} can; we have no choice but to write it as an unevaluated root (a surd). For simplicity and economy, we usually represent this number by the symbol \displaystyle i (sometimes \displaystyle j). It is sometimes called the imaginary unit, and any number of the form \displaystyle b\,i, where \displaystyle b is real, is known as an imaginary number. We go on to define a complex number as an object that can be written in the form

\displaystyle z=a+bi\,\mbox{,}

where \displaystyle a and \displaystyle b are real numbers, and \displaystyle i satisfies \displaystyle i^2=-1.

If \displaystyle a = 0 then the number is "purely imaginary". If \displaystyle b = 0 the number is real. We can see that the real numbers are a subset of the complex numbers. The set of complex numbers is designated by C.

For an arbitrary complex number one often uses the symbol \displaystyle z. If \displaystyle z=a+bi, where \displaystyle a and \displaystyle b are real, then \displaystyle a is the real part and \displaystyle b the imaginary part of \displaystyle z. One uses the following notation:

\displaystyle \begin{align*}a &= \mathop{\rm Re} z\,\mbox{,}\\ b&=\mathop{\rm Im} z\,\mbox{.}\end{align*}

When one calculates with complex numbers one treats them just like real numbers, but keeps track of the fact that \displaystyle i^2=-1.


Addition and subtraction

To add or subtract complex numbers one adds or subtracts the real and imaginary parts separately. If \displaystyle z=a+bi and \displaystyle w=c+di are two complex numbers then,

\displaystyle \begin{align*} z+w &= a+bi+c+di = a+c+(b+d)i\,\mbox{,}\\ z-w &= a+bi-(c+di) = a-c+(b-d)i\,\mbox{.}\end{align*}

Example 2

  1. \displaystyle (3-5i)+(-4+i)=-1-4i
  2. \displaystyle \bigl(\tfrac{1}{2}+2i\bigr)-\bigl(\tfrac{1}{6}+3i\bigr) = \tfrac{1}{3}-i
  3. \displaystyle \frac{3+2i}{5}-\frac{3-i}{2} = \frac{6+4i}{10}-\frac{15-5i}{10} = \frac{-9+9i}{10} = -0\textrm{.}9 + 0\textrm{.}9i


Multiplication

Complex numbers are multiplied in the same way as ordinary real numbers or algebraic expressions, with the extra condition that \displaystyle i^2=-1. Generally one has for two complex numbers \displaystyle z=a+bi and \displaystyle w=c+di that

\displaystyle z\, w=(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i\,\mbox{.}

Example 3

  1. \displaystyle 3(4-i)=12-3i
  2. \displaystyle 2i(3-5i)=6i-10i^2=10+6i
  3. \displaystyle (1+i)(2+3i)=2+3i+2i+3i^2=-1+5i
  4. \displaystyle (3+2i)(3-2i)=3^2-(2i)^2=9-4i^2=13
  5. \displaystyle (3+i)^2=3^2+2\times3i+i^2=8+6i
  6. \displaystyle i^{12}=(i^2)^6=(-1)^6=1
  7. \displaystyle i^{23}=i^{22}\times i=(i^2)^{11}\times i=(-1)^{11}i=-i


Complex conjugate

If \displaystyle z=a+bi then \displaystyle \overline{z} = a-bi is called the complex conjugate of \displaystyle z (the opposite is also true, that \displaystyle z is conjugate to \displaystyle \overline{z}). One obtains the relationships

\displaystyle \begin{align*} z+\overline{z} &= a + bi + a - bi = 2a = 2\, \mathop{\rm Re} z\,\mbox{,}\\ z-\overline{z} &= a + bi - (a - bi) = 2bi = 2i\, \mathop{\rm Im} z\,\mbox{,}\end{align*}

but most importantly, using the difference of two squares rule, one obtains

\displaystyle z\, \bar z = (a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2i^2=a^2+b^2\,\mbox{,}

i.e. that the product of a complex number and its conjugate is always real.


Example 4

  1. \displaystyle z=5+i\qquad then \displaystyle \quad\overline{z}=5-i\,.
  2. \displaystyle z=-3-2i\qquad then \displaystyle \quad\overline{z} =-3+2i\,.
  3. \displaystyle z=17\qquad then \displaystyle \quad\overline{z} =17\,.
  4. \displaystyle z=i\qquad then \displaystyle \quad\overline{z} =-i\,.
  5. \displaystyle z=-5i\qquad then \displaystyle \quad\overline{z} =5i\,.

Example 5

  1. If \displaystyle z=4+3i one has
    • \displaystyle z+\overline{z} = 4 + 3i + 4 -3i = 8
    • \displaystyle z-\overline{z} = 6i
    • \displaystyle z \, \overline{z} = 4^2-(3i)^2=16+9=25
  2. If for \displaystyle z one has \displaystyle \mathop{\rm Re} z=-2 and \displaystyle \mathop{\rm Im} z=1, one gets
    • \displaystyle z+\overline{z} = 2\,\mathop{\rm Re} z = -4
    • \displaystyle z-\overline{z} = 2i\,\mathop{\rm Im} z = 2i
    • \displaystyle z\, \overline{z} = (-2)^2+1^2=5


Division

For the division of two complex numbers one multiplies the numerator and denominator by the complex conjugate of the denominator, thus getting a denominator which is a real number. Then, both the real and imaginary parts of the (new) numerator are divided by this number (the new denominator). In general, if \displaystyle z=a+bi and \displaystyle w=c+di:

\displaystyle \frac{z}{w} = \frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}\,i

Example 6

  1. \displaystyle \quad\frac{4+2i}{1+i} = \frac{(4+2i)(1-i)}{(1+i)(1-i)} = \frac{4-4i+2i-2i^2}{1-i^2} = \frac{6-2i}{2}=3-i
  2. \displaystyle \quad\frac{25}{3-4i} = \frac{25(3+4i)}{(3-4i)(3+4i)} = \frac{25(3+4i)}{3^2-16i^2} = \frac{25(3+4i)}{25} = 3+4i
  3. \displaystyle \quad\frac{3-2i}{i} = \frac{(3-2i)(-i)}{i(-i)} = \frac{-3i+2i^2}{-i^2} = \frac{-2-3i}{1} = -2-3i

Example 7

  1. \displaystyle \quad\frac{2}{2-i}-\frac{i}{1+i} = \frac{2(2+i)}{(2-i)(2+i)} - \frac{i(1-i)}{(1+i)(1-i)} = \frac{4+2i}{5}-\frac{1+i}{2}
    \displaystyle \quad\phantom{\frac{2}{2-i}-\frac{i}{1+i}}{} = \frac{8+4i}{10}-\frac{5+5i}{10} = \frac{3-i}{10}\vphantom{\Biggl(}
  2. \displaystyle \quad\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}} = \frac{\dfrac{1-i}{1-i}-\dfrac{2}{1-i}}{\dfrac{2i(2+i)}{(2+i)} + \dfrac{i}{2+i}} = \frac{\dfrac{1-i-2}{1-i}}{\dfrac{4i+2i^2 + i}{2+i}} = \frac{\dfrac{-1-i}{1-i}}{\dfrac{-2+5i}{2+i}}
    \displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-1-i}{1-i}\times \frac{2+i}{-2+5i} = \frac{(-1-i)(2+i)}{(1-i)(-2+5i)} = \frac{-2-i-2i-i^2}{-2+5i+2i-5i^2}\vphantom{\Biggl(}
    \displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-1-3i}{3+7i} = \frac{(-1-3i)(3-7i)}{(3+7i)(3-7i)} = \frac{-3+7i-9i+21i^2}{3^2-49i^2}\vphantom{\Biggl(} \displaystyle \quad\phantom{\smash{\frac{1-\dfrac{2}{1-i}}{2i+\dfrac{i}{2+i}}}}{} = \frac{-24-2i}{58} = \frac{-12-i}{29}\vphantom{\Biggl(}

Example 8

Determine the real number \displaystyle a such that the expression \displaystyle \ \frac{2-3i}{2+ai}\ becomes real.

Multiply the numerator and denominator by the complex conjugate of the denominator so that the expression can be written with separate real and imaginary parts.

\displaystyle \frac{(2-3i)(2-ai)}{(2+ai)(2-ai)} = \frac{4-2ai-6i+3ai^2}{4-a^2i^2} = \frac{4-3a-(2a+6)i}{4+a^2}

If the expression is to be real , the imaginary part must be 0, ie.

\displaystyle 2a+6=0\quad\Leftrightarrow\quad a = -3\,\mbox{.}


Equations

If the two complex numbers \displaystyle z=a+bi and \displaystyle w=c+di are equal then it is not hard to show that both the real and imaginary parts must be equal (in other words, \displaystyle a=c and \displaystyle b=d). When you are looking for an unknown complex number \displaystyle z in an equation, you can either try to solve for the number \displaystyle z in the usual way, or insert \displaystyle z=a+bi in the equation and then compare the real and imaginary parts of the two sides of the equation with each other.

Example 9

  1. Solve the equation \displaystyle 3z+1-i=z-3+7i.

    Collect all \displaystyle z on the left-hand side by subtracting \displaystyle z from both sides
    \displaystyle 2z+1-i = -3+7i
    and now subtract \displaystyle 1-i
    \displaystyle 2z = -4+8i\,\mbox{.}
    This gives that \displaystyle \ z=\frac{-4+8i}{2}=-2+4i\,\mbox{.}
  2. Solve the equation \displaystyle z(-1-i)=6-2i.

    Divide both sides by \displaystyle -1-i in order to obtain \displaystyle z
    \displaystyle z = \frac{6-2i}{-1-i} = \frac{(6-2i)(-1+i)}{(-1-i)(-1+i)} = \frac{-6+6i+2i-2i^2}{(-1)^2 -i^2} = \frac{-4+8i}{2} = -2+4i\,\mbox{.}
  3. Solve the equation \displaystyle 3iz-2i=1-z.

    Adding \displaystyle z and \displaystyle 2i to both sides gives
    \displaystyle 3iz+z=1+2i\quad \Leftrightarrow \quad z(3i+1)=1+2i\,\mbox{.}

    This gives

    \displaystyle z = \frac{1+2i}{1+3i} = \frac{(1+2i)(1-3i)}{(1+3i)(1-3i)} = \frac{1-3i+2i-6i^2}{1-9i^2} = \frac{7-i}{10}\,\mbox{.}
  4. Solve the equation \displaystyle 2z+1-i=\bar z +3 + 2i.

    The equation contains both \displaystyle z as well as \displaystyle \overline{z} and therefore we assume \displaystyle z to be \displaystyle z=a+ib and solve the equation for \displaystyle a and \displaystyle b by equating the real and imaginary parts of both sides
    \displaystyle 2(a+bi)+1-i=(a-bi)+3+2i

    i.e.

    \displaystyle (2a+1)+(2b-1)i=(a+3)+(2-b)i\,\mbox{,}

    which gives

    \displaystyle \left\{\begin{align*} 2a+1 &= a+3\\ 2b-1 &= 2-b\end{align*}\right.\quad\Leftrightarrow\quad\left\{\begin{align*} a &= 2\\ b &= 1\end{align*}\right.\,\mbox{.}
    The answer is therefore, \displaystyle z=2+i.


Study advice

Keep in mind that:

Calculations with complex numbers are done in the same way as with ordinary numbers with the additional information that \displaystyle i^2=-1.

Quotients of complex numbers are simplified by multiplying the numerator and denominator by the complex conjugate of the denominator.

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