3.1 Rechnungen mit komplexen Zahlen

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{{Vald flik|[[3.1 Räkning med komplexa tal|Teori]]}}
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{{Vald flik|[[3.1 Räkning med komplexa tal|Theory]]}}
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{{Ej vald flik|[[3.1 Övningar|Övningar]]}}
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{{Ej vald flik|[[3.1 Övningar|Exercises]]}}
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'''Innehåll:'''
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'''Content:'''
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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
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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 real numbers are not enough to be solutions of all possible algebraic equations, in other words. there are equations of the type
{{Fristående formel||<math>a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0</math>}}
{{Fristående formel||<math>a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0</math>}}
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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.
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which 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 solves the equation <math>x^2=-1</math> and manipulate <math>\sqrt{-1}</math> as any other number, it turns out that every algebraic equation has solutions.
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The number <math>\sqrt{-1}</math> however is not real number, we cannot go out into the world and measure <math>\sqrt{-1}</math> anywhere, or find something that is numerically <math>\sqrt{-1}</math>, but we can still have great use of this number in many real contexts.
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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.
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<div class="exempel">
<div class="exempel">
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'''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 the non-real number <math>\sqrt{-1}</math>. If we allows 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 certainly is a 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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In order to solve our problem we had to use a number that was not real in order to arrive at the real number solution.
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== Definition av komplexa tal ==
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== Definition of complex numbers==
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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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One introduces the ''imaginary unit'' <math>i=\sqrt{-1}</math> and define a ''complex number'' as an object that can be written in the form
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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>.
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where <math>a</math> and <math>b</math> are real numbers, and <math>i</math> satisfy <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 beteckningen <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\>
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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 in principle 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 ==
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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 (subtracts) the real and imaginary parts separately.
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If <math>z=a+bi</math> and <math>w=c+di</math> are two complex numbers then,
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'''Exempel 2'''
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''' Example 2'''
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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
<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>}}
<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="exempel">
<div class="exempel">
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'''Exempel 3'''
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''' Example 3'''
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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
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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
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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.
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<div class="exempel">
<div class="exempel">
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'''Exempel 4'''
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''' Example 4'''
<ol type="a">
<ol type="a">
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<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">
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'''Exempel 5'''
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''' Example 5'''
<ol type="a">
<ol type="a">
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<li> Om <math>z=4+3i</math> då gäller att
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<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>
*<math>z \cdot \overline{z} = 4^2-(3i)^2=16+9=25</math>
*<math>z \cdot \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
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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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== 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>:
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For the division of two complex numbers one multiplies the numerator and denominator with the complex conjugate of the denominator thus getting a denominator which is a real number. Thereafter, both the real and imaginary parts of the numerator is divided by this number (the new denominator). In general, if <math>z=a+bi</math> and <math>w=c+di</math>:
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'''Exempel 6'''
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''' Example 6'''
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'''Exempel 7'''
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''' Example 7'''
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<div class="exempel">
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'''Exempel 8'''
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''' Example 8'''
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Bestäm det reella talet <math>a</math> så att uttrycket <math>\ \frac{2-3i}{2+ai}\ </math> blir reellt.
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Determine the real number <math>a</math> so that the expression <math>\ \frac{2-3i}{2+ai}\ </math> becomes real.
<br\>
<br\>
<br\>
<br\>
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Förläng med nämnarens konjugat så att uttrycket kan skrivas med separata real- och imaginärdelar
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Multiply the numerator and denominator with 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>}}
{{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>}}
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Om uttrycket ska bli reellt så måste imaginärdelen vara 0, dvs.
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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>}}
{{Fristående formel||<math>2a+6=0\quad\Leftrightarrow\quad a = -3\,\mbox{.}</math>}}
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== Ekvationer ==
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== 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.
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For two complex numbers <math>z=a+bi</math> and <math>w=c+di</math> to be equal, requires that both the real and imaginary parts are equal, in other words. that <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">
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'''Exempel 9'''
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''' Example 9'''
<ol type="a">
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<li> Lös ekvationen <math>3z+1-i=z-3+7i</math>.
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<li> Solve the equation <math>3z+1-i=z-3+7i</math>.
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<br/>
<br/>
<br/>
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Samla <math>z</math> i vänsterledet genom att subtrahera båda led med <math>z</math>
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Collect all <math>z</math> on the left-hand side by subtracting <math>z</math> from both sides
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{{Fristående formel||<math>2z+1-i = -3+7i</math>}}
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{{Fristående formel||<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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{{Fristående formel||<math>2z = -4+8i\,\mbox{.}</math>}}
{{Fristående formel||<math>2z = -4+8i\,\mbox{.}</math>}}
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Detta ger att <math>\ z=\frac{-4+8i}{2}=-2+4i\,\mbox{.}</math></li>
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This gives that <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>
{{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>
{{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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<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
{{Fristående formel||<math>3iz+z=1+2i\quad \Leftrightarrow \quad z(3i+1)=1+2i\,\mbox{.}</math>}}
{{Fristående formel||<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
{{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>
{{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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<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 {{Fristående formel||<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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{{Fristående formel||<math>(2a+1)+(2b-1)i=(a+3)+(2-b)i\,\mbox{,}</math>}}
{{Fristående formel||<math>(2a+1)+(2b-1)i=(a+3)+(2-b)i\,\mbox{,}</math>}}
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vilket ger att
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which gives
{{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>}}
{{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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Svaret är alltså <math>z=2+i</math>.</li>
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The answer is therefore, <math>z=2+i</math>.</li>
</ol>
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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>.
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Quotients of complex numbers are simplified by multiplying the numerator and denominator with the complex conjugate of the denominator.
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Version vom 08:00, 23. Jul. 2008

 
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Content:

  • 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 real numbers are not enough to be solutions of all possible algebraic equations, in other words. there are equations of the type

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which 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 solves the equation \displaystyle x^2=-1 and manipulate \displaystyle \sqrt{-1} as any other number, it turns out that every algebraic equation has solutions. The number \displaystyle \sqrt{-1} however is not real number, we cannot go out into the world and measure \displaystyle \sqrt{-1} anywhere, or find something that is numerically \displaystyle \sqrt{-1}, but we can still have great use of this number in many real contexts.


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 the non-real number \displaystyle \sqrt{-1}. If we allows 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 certainly is a real number.

In order to solve our problem we had to use a number that was not real in order to arrive at the real number solution.


Definition of complex numbers

One introduces the imaginary unit \displaystyle i=\sqrt{-1} and define a complex number as an object that can be written in the form

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where \displaystyle a and \displaystyle b are real numbers, and \displaystyle i satisfy \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 beteckningen \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:

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When one calculates with complex numbers one treats them in principle 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 (subtracts) the real and imaginary parts separately. If \displaystyle z=a+bi and \displaystyle w=c+di are two complex numbers then,

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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{,}9 + 0{,}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

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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\cdot3i+i^2=8+6i
  6. \displaystyle i^{12}=(i^2)^6=(-1)^6=1
  7. \displaystyle i^{23}=i^{22}\cdot i=(i^2)^{11}\cdot 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

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

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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 \cdot \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\cdot \overline{z} = (-2)^2+1^2=5


Division

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

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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}\cdot \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 so that the expression \displaystyle \ \frac{2-3i}{2+ai}\ becomes real.

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

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

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Equations

For two complex numbers \displaystyle z=a+bi and \displaystyle w=c+di to be equal, requires that both the real and imaginary parts are equal, in other words. that \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
    1. REDIRECT Template:Abgesetzte Formel and now subtract \displaystyle 1-i
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    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
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  3. Solve the equation \displaystyle 3iz-2i=1-z.

    Adding \displaystyle z and \displaystyle 2i to both sides gives
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    This gives
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  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
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    i.e.
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    which gives
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    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 with the complex conjugate of the denominator.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/3.1_Rechnungen_mit_komplexen_Zahlen“