1
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Ok, and how is their divisor then imaginary? $\endgroup$ – TestGuest Nov 20 '13 at 11:54
  • 2
    $\begingroup$ 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 $\operatorname{Re} z\overline{w} = 0$. $\endgroup$ – Daniel Fischer Nov 20 '13 at 11:56
  • $\begingroup$ 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)? $\endgroup$ – TestGuest Nov 20 '13 at 16:47
  • 1
    $\begingroup$ @TestGuest Yes. Is it also called rhombus in English? Didn't know that, thanks for the heads-up. $\endgroup$ – Daniel Fischer Nov 20 '13 at 16:53
2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ I like this! +1 $\endgroup$ – Namaste Nov 20 '13 at 14:57
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.