### Complex numbers: Calculating with complex numbers

### Geometric interpretation

Let #z# and #w# be two complex numbers. The absolute value #|z -w|# has as geometric interpretation the distance from #z# to #w#. After all, #|z - w| = \sqrt{(\Re(z)-\Re(w))^2 + (\Im(z)-\Im(w))^2}#, while the *Pythagorean theorem* shows us that this is the distance from #z# to #w#. We use this interpretation to describe the circle in the flat plane with a complex equation.

Circles and lines in terms of polar coordinates

Let #a# and #b# be real numbers and write #w=a+b\cdot\ii#.

- Let #r# be a non-negative real number. The set of complex numbers #z=x+y\cdot\ii#, with real #x# and #y#, that together form a circle of radius #r# and center #\rv{a,b}#, is not only given by the real equation \[(x-a)^2+(y-b)^2=r^2\tiny,\] but also by the complex equation \[\left|z-w\right|=r\tiny.\]
- Let #\varphi# be a real number. The set of complex numbers #z=x+y\cdot\ii#, with real #x# and #y# that together form the line through #\rv{a,b}# with
*slope*#\tan(\varphi)#, is not only given by the real equation \[\sin(\varphi)\cdot(x-a)-\cos(\varphi)\cdot(y-b) =0\tiny,\] but also by #z=w# and the solutions to the complex equation \[\arg(z-w)=\varphi\pmod{\pi}\tiny.\]

We say that the circle is **defined **by the equation #\left|z-w\right|=r# with unknown #z# and that the line is **defined** by #\arg(z-w)=\varphi\pmod{\pi}#.

We assume that the reductions of the real equations are known and we focus on the complex ones:

1. The equation #|z-(a+b\cdot\ii) | = r# with unknown #z# has as solutions exactly those complex numbers #z# with distance #r# to #a+b\cdot\ii#. This is a circle with center #\rv{a,b}# and radius #r#.

2. The equation #\arg(z-w)=\varphi\pmod{\pi}# with unknown #z# has as solutions the complex numbers of the form #z=w+\lambda\cdot\left(\cos(\varphi)+\sin(\varphi)\cdot\ii\right)#, in which #\lambda# is a real number unequal to #0#. But #\lambda=0# also occurs because we allow #z=w#. Hence, we get the line through #w# with direction vector #\rv{\cos(\varphi),\sin(\varphi)}# and thus with slope #\tan(\varphi)#.

The reductions of the corresponding real equations are known from the theory for *circles,* respectively *lines,* in the plane.

Notice the choice for #\pmod{\pi}# instead of #\pmod{2\pi}#. This neglects the direction of the line, which is important to the polar coordinates to describe points in a unique way.

If #\varphi=\pm\frac{\pi}{2}#, then #\tan(\varphi)# is undefined and neither is the slope. However, the direction vector then is #\rv{\pm1,0}#, not posing a problem (except that the line is not a graph of a linear function).

For the line there is also a similar result without #\arg(z)#: if #a#, #b#, and #c# are real numbers with #a# and/or #b# unequal to #0#, then #a\cdot x+b\cdot y+c=0# is the equation of a line. The set of complex numbers #z=x+y\cdot\ii#, with real #x# and #y#, which forms this line, can be described by the equation #a\cdot\Re(z) +b\cdot\Im(z)+c=0#, or, more compactly, by #\Re\left(\left(a-b\,\ii\right)\cdot z\right)+c=0#.

The equation #|z - a| = |z-b|# has as solutions all points #z# that have the same distance to #a# and #b#. This is the *perpendicular line* between #a# and #b#. Other objects in the flat plane, such as ellipses and perpendicular lines, can be expressed in complex formulas, as will be shown in some exercises.

The radius is \(2\)

If \(z=x+y\,\ii\) is the standard form of #z#, then the equation \[|z+0|=2\] means, according to the definition of

*the absolute value of a complex number*\[x^2+y^2=4\] This is exactly the equation of a circle with center \(\rv{0,0}\) and radius \(2\).

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.