# Geometric Solution for Equation with Complex Numbers

Given we have two complex numbers $z_1$ and $z_2$ with $|z_1| = |z_2|$.

How can it be shown geometrically, that $\frac{z_1+z_2}{z_1-z_2}$ is purely imaginery?

Consider the quadrilateral with vertices $0, z_1, z_1+z_2, z_2$. The condition $\lvert z_1\rvert = \lvert z_2 \rvert$ makes it a rhombus. Thus the two diagonals $z_1+z_2$ and $z_1 - z_2$ are orthogonal.

• Ok, and how is their divisor then imaginary? – TestGuest Nov 20 '13 at 11:54
• Two (nonzero) complex numbers are orthogonal to each other if and only if their quotient is purely imaginary. You can see it via $\frac{z}{w} = \frac{z\overline{w}}{\lvert w\rvert^2}$ if you're more familiar with the characterisation of orthogonality via $\operatorname{Re} z\overline{w} = 0$. – Daniel Fischer Nov 20 '13 at 11:56
• All complex numbers with the same abs. value can be imagined as being on a circle with radius equal the absolute value of those, right? Btw, I think you mean rhombus (equal sides)? – TestGuest Nov 20 '13 at 16:47
• @TestGuest Yes. Is it also called rhombus in English? Didn't know that, thanks for the heads-up. – Daniel Fischer Nov 20 '13 at 16:53

Here is an elementary way for checking that we don't have a real part for your complex fraction. Let we have $$z=\frac{z_1}{z_2}=\frac{x_1+iy_1}{x_2+iy_2}$$ then with a tedious hand calculation we get: $$z=\left(\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}\right)+i\left(\frac{y_1x_2-y_2x_1}{x_2^2+y_2^2}\right)$$ This is for $z_1$ and $z_2$. Now you set $Z_1=z_1+z_2$ and $Z_2=z_1-z_2$ in above result and by assuming $$x_1^2-x_2^2=-(y_1^2+y_2^2)$$ you'll get the final result. Sorry if this way is not a geometrically one.

• I like this! +1 – Namaste Nov 20 '13 at 14:57

Firstly $z_1$ and $z_2$ lie on a circle of radius $r=|z_1|=|z_2|$. Now wherever $A=z_1+z_2$ is, $B=z_1-z_2=z_1+z_2-(2z_2)$ is found by reflecting $z_1+z_2$ through the point $z_1$.

Now consider the circle with centre $z_1$ and radius $r$ (and so contains $A$ and $B$ - draw a picture). Now because $|z_1|=r$ we have that $\angle AOB$ is the angle in a semicircle and so is $\pi/2$.

Therefore $$\arg A=\arg B\pm\pi/2\Rightarrow \arg A-\arg B=\pm\pi/2.$$

Now define the multiplication of complex numbers $z$ by $w$ geometrically: then $z\times w$ is $z$ is stretched by a factor $|w|$ and rotated through an angle $\arg w$ in the anti-clockwise sense. If you don't want to do that just show $$(r(\cos\theta+i\sin\theta))(s(\cos\alpha)+i\sin\alpha)=rs(\cos(\theta+\alpha)+i\sin(\theta+\alpha)).$$

Dividing by $w$ includes a clockwise rotation through an angle $\arg w$ - i.e. taking $\arg w$ away from $\arg z$.

So we have $$\frac{z_1+z_2}{z_1-z_2}=\frac{A}{B}.$$ Now starting at $A$, dividing by $B$ includes a clockwise rotation through an angle $\arg B$ so $$\arg\frac{A}{B}=\arg A-\arg B,$$ and we have already shown that this is $\pm\pi/2$. Therefore $\displaystyle\frac{z_1+z_2}{z_1-z_2}$ is imaginary.