Complex numbers: Calculating with complex numbers
Calculating with polar coordinates
Complex numbers in standard form are easy to add, but multiplication is a bit more complicated. If we use polar coordinates, the situation is the other way around: addition is not obvious, but multiplication has become easy.
Multiplication in polar coordinates
The multiplication of two complex numbers #z_1# and #z_2# can be described as follows in terms of their abolute values #|z_1 |# and #|z_2|# and their arguments #\varphi_1# and #\varphi_2#.
\[
\begin{array}{rcl}
|z_1\cdot z_2|&=& |z_1| \cdot |z_2|,\\
\arg(z_1\cdot z_2) & = &\arg (z_1)+\arg(z_2) \pmod{2\pi}
\end {array}
\]
By repeated application of these rules, for #n# we find complex numbers #z_1,z_2, \ldots ,z_n#
\[
\begin{array}{rcl}|z_1\cdot z_2 \cdots z_n |&=& |z_1|\cdot \ |z_2|\cdots |z_n |,\\
\arg(z_1\cdot z_2 \cdots z_n ) & =& \arg (z_1)+\arg(z_2) + \cdots + \arg(z_n)
\, \pmod{2\pi}\end {array}
\]
When calculating with complex numbers, you can always choose whether you want to use the representation in Cartesian coordinates or in polar coordinates. The following rules of thumb can be applied:
- if the calculation mainly contains additions, then using Cartesian coordinates is advisable;
- if the calculation mainly contains multiplications, then polar coordinates is the preferred option.
When solving equations it is important that:
- two complex numbers in standard form are equal to one another if and only if both the real and imaginary parts are equal to each other.
- two complex numbers unequal to #0# are equal to one another if and only if they have the same absolute values and the arguments are equal modulo #2 \pi#.
With these properties, we can convert complex equations into real equations. Note that this is not always convenient or necessary.
\[
x^2 +2x\cdot y\cdot\ii -y^2 = 2 \ii\tiny.
\]
Equating the real and imaginary parts, gives the two equations
\[\eqs{x^2-y^2 &=&0\cr 2x\cdot y &=& 2}\] From the first equation with the real unknowns #x# and #y# follows #x =y# or #x=-y#. Entering this in the second one gives #x^2 =1# and #x^2=-1#. The equation #x^2 =-1# does not have (real!) solutions, and the equation #x^2 =1# leads to #x=1# or #x=-1#. If we translate these results to #z=x+y\cdot\ii#, we find the solution \[z=1 +\ii\,\lor \,z=-1-\ii\tiny,\] where #\lor# is the logical "or".
Second solution: Using the absolute value and the argument we reason as follows.
\[ \begin{array}{rcl}
z^2 \phantom{x}&=& \phantom{x}2\,\ii \\
|z^2| =2 \phantom{x} &\text{and}& \phantom{x}
\arg(z^2)=\frac{\pi}{2} +2k\cdot\pi \phantom{x} (k\text{ integer}) \\
|z|^2 = 2 \phantom{x}&\text{and}& \phantom{x}2\arg(z) =\frac{\pi}{2} +2k\cdot\pi\phantom{x} (k\text{ integer})\\
|z| = \sqrt{2} \phantom{x} &\text{and}&\phantom{x}\arg(z) =\frac{\pi}{4} +k\cdot\pi \phantom{x}(k\text{ integer})
\end {array}
\]
From this follows the solution:
\[
z=\sqrt{2}\, \left(\cos \left(\frac{\pi}{4}\right) +\sin \left(\frac{\pi}{4}\right)\cdot\ii\right)
\, \lor \, z=
\sqrt{2}\, \left(\cos \left (\frac{5\pi}{4}\right) +\sin \left(\frac{5\pi}{4}\right)\cdot\ii\right)\tiny.
\]
Here we only use the values #k=0# and #k=1#, because for #k=2# we again find the same solution as for #k=0# and for #k=3# we again find the same solution as for #k=1#, and so on.
Verify that the two complex numbers are the same as described in the first solution.
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