3.1 Rechnungen mit komplexen Zahlen
Aus Online Mathematik Brückenkurs 2
| 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 real numbers are not enough to be 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 |
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 a 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 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 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
| \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 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,
| \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
- \displaystyle (3-5i)+(-4+i)=-1-4i
- \displaystyle \bigl(\tfrac{1}{2}+2i\bigr)-\bigl(\tfrac{1}{6}+3i\bigr) = \tfrac{1}{3}-i
- \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\cdot w=(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(ad+bc)i\,\mbox{.} |
Example 3
- \displaystyle 3(4-i)=12-3i
- \displaystyle 2i(3-5i)=6i-10i^2=10+6i
- \displaystyle (1+i)(2+3i)=2+3i+2i+3i^2=-1+5i
- \displaystyle (3+2i)(3-2i)=3^2-(2i)^2=9-4i^2=13
- \displaystyle (3+i)^2=3^2+2\cdot3i+i^2=8+6i
- \displaystyle i^{12}=(i^2)^6=(-1)^6=1
- \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
| \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\cdot \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
- \displaystyle z=5+i\qquad then \displaystyle \quad\overline{z}=5-i\,.
- \displaystyle z=-3-2i\qquad then \displaystyle \quad\overline{z} =-3+2i\,.
- \displaystyle z=17\qquad then \displaystyle \quad\overline{z} =17\,.
- \displaystyle z=i\qquad then \displaystyle \quad\overline{z} =-i\,.
- \displaystyle z=-5i\qquad then \displaystyle \quad\overline{z} =5i\,.
Example 5
- 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
- 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:
| \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
- \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
- \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
- \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
- \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(} - \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.
| \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
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
- 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
and now subtract \displaystyle 1-i\displaystyle 2z+1-i = -3+7i
This gives that \displaystyle \ z=\frac{-4+8i}{2}=-2+4i\,\mbox{.}\displaystyle 2z = -4+8i\,\mbox{.} - 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{.} - 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{.} - 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
The answer is therefore, \displaystyle z=2+i.\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{.}
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.
