3.1 Rechnungen mit komplexen Zahlen
Aus Online Mathematik Brückenkurs 2
Theorie | Übungen |
Inhalt:
- Real and imaginary part
- Addition and subtraction of complex numbers
- Complex conjugate
- Multiplication and division of complex numbers
Lernziele:
Nach diesem Abschnitt sollst Du folgendes können 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
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that do not have a solution among the real numbers. For example, the equation −1
−1
−1
−1
−1
Beispiel 1
If we would like to find out the sum of the roots (solutions) of the equation −1
−1
−1
−1
−1+1−
−1=2
Even though the answer to the problem was a "real" number, we found the "imaginary" number −1
Definition of complex numbers
The terms "real" (for the ordinary, familiar positive and negative numbers, together with zero) and "imaginary" (for numbers like −1
−1
The number −1
2
where
If
For an arbitrary complex number one often uses the symbol
When one calculates with complex numbers one treats them just like real numbers, but keeps track of the fact that
Addition and subtraction
To add or subtract complex numbers one adds or subtracts the real and imaginary parts separately.
If
Beispiel 2
-
(3−5i)+(−4+i)=−1−4i -
21+2i
−
61+3i
=31−i
-
53+2i−23−i=106+4i−1015−5i=10−9+9i=−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
Beispiel 3
3(4−i)=12−3i 2i(3−5i)=6i−10i2=10+6i (1+i)(2+3i)=2+3i+2i+3i2=−1+5i (3+2i)(3−2i)=32−(2i)2=9−4i2=13 (3+i)2=32+2 3i+i2=8+6i
i12=(i2)6=(−1)6=1 i23=i22 i=(i2)11
i=(−1)11i=−i
Complex conjugate
If
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.
Beispiel 4
z=5+i thenz=5−i .z=−3−2i thenz=−3+2i .z=17 thenz=17 .z=i thenz=−i .z=−5i thenz=5i .
Beispiel 5
- If
z=4+3i one hasz+z=4+3i+4−3i=8 z−z=6i zz=42−(3i)2=16+9=25
- If for
z one hasRez=−2 andImz=1 , one getsz+z=2Rez=−4 z−z=2iImz=2i zz=(−2)2+12=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
Beispiel 6
1+i4+2i=(1+i)(1−i)(4+2i)(1−i)=1−i24−4i+2i−2i2=26−2i=3−i 253−4i=25(3+4i)(3−4i)(3+4i)=32−16i225(3+4i)=2525(3+4i)=3+4i i3−2i=i(−i)(3−2i)(−i)=−i2−3i+2i2=1−2−3i=−2−3i
Beispiel 7
22−i−i1+i=2(2+i)(2−i)(2+i)−i(1−i)(1+i)(1−i)=54+2i−21+i
=108+4i−105+5i=103−i 1−21−i2i+i2+i=1−i1−i−21−i(2+i)2i(2+i)+i2+i=1−i1−i−22+i4i+2i2+i=1−i−1−i2+i−2+5i
=1−i−1−i 2+i−2+5i=(−1−i)(2+i)(1−i)(−2+5i)=−2−i−2i−i2−2+5i+2i−5i2
=3+7i−1−3i=(3+7i)(3−7i)(−1−3i)(3−7i)=32−49i2−3+7i−9i+21i2 =58−24−2i=29−12−i
Beispiel 8
Determine the real number
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.
If the expression is to be real , the imaginary part must be 0, ie.
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Equations
If the two complex numbers
Beispiel 9
- Solve the equation
3z+1−i=z−3+7i .
Collect allz on the left-hand side by subtractingz from both sides2z+1−i=−3+7i 1−i 2z=−4+8i. z=2−4+8i=−2+4i. - Solve the equation
z(−1−i)=6−2i .
Divide both sides by−1−i in order to obtainz z=6−2i−1−i=(6−2i)(−1+i)(−1−i)(−1+i)=(−1)2−i2−6+6i+2i−2i2=2−4+8i=−2+4i. - Solve the equation
3iz−2i=1−z .
Addingz and2i to both sides gives3iz+z=1+2i z(3i+1)=1+2i.
This gives
z=1+3i1+2i=(1+3i)(1−3i)(1+2i)(1−3i)=1−9i21−3i+2i−6i2=107−i. - Solve the equation
2z+1−i=z .+3+2i
The equation contains bothz as well asz and therefore we assumez to bez=a+ib and solve the equation fora andb by equating the real and imaginary parts of both sides2(a+bi)+1−i=(a−bi)+3+2i i.e.
(2a+1)+(2b−1)i=(a+3)+(2−b)i, which gives
2a+12b−1=a+3=2−b
ab=2=1.
z=2+i .
Tipps fürs lernen
Keep in mind that:
Calculations with complex numbers are done in the same way as with ordinary numbers with the additional information that
Quotients of complex numbers are simplified by multiplying the numerator and denominator by the complex conjugate of the denominator.