3.1 Rechnungen mit komplexen Zahlen
Aus Online Mathematik Brückenkurs 2
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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
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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 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
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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:
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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
- \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
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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
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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
- \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:
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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.
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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
- 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- REDIRECT Template:Abgesetzte Formel and now subtract \displaystyle 1-i
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- Solve the equation \displaystyle z(-1-i)=6-2i.
Divide both sides by \displaystyle -1-i in order to obtain \displaystyle z- REDIRECT Template:Abgesetzte Formel
Adding \displaystyle z and \displaystyle 2i to both sides gives
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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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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.
