20
$\begingroup$

If $$\left|\frac{Z_1 - iZ_2}{Z_1 + iZ_2}\right| = 1$$ then prove that $Z_1/Z_2$ is real .

This is how I proceeded.

Dividing throughout by $Z_2$ we will have

$$\left|{\frac{\frac{Z_1}{Z_2} - i}{\frac{Z_1}{Z_2} + i}}\right| = 1$$

Thus

$$\left|\frac{Z_1}{Z_2} - i\right|= \left|\frac{Z_1}{Z_2} + i\right|$$

How do proceed from here ?

According to the solution of the above problem , the previous statement would imply that $\frac{Z_1}{Z_2}$ is equidistant from $i$ and $-i$. Thus it is real. Now how does that make it real ? Please help me with this.

$\endgroup$
  • 5
    $\begingroup$ If you plot $i$ and $-i$ on coordinate plane it would be $(0,1)$ and $(0,-1)$ respectively. $\frac{Z_1}{Z_2}$ is at equal distance from both these points. So It has to be on X-axis. Hence it has to be real. $\endgroup$ – MeetR Jan 16 '18 at 17:46
  • 6
    $\begingroup$ The statement is not true for $Z_2=0$ (in that case, $Z_1/Z_2$ is not even defined) $\endgroup$ – chtz Jan 17 '18 at 9:23
  • 1
    $\begingroup$ @chtz: If $Z_2 = 0$, then the problem is ill-posed. $\endgroup$ – Kevin Jan 18 '18 at 4:23
  • 1
    $\begingroup$ @Kevin Yes, and since that case was not excluded it makes the statement false. An example of a true statement would have been: "If $|Z_1-iZ_2|=|Z_1+iZ_2|$ then $Z_1 \cdot \bar{Z}_2$ is real." $\endgroup$ – chtz Jan 18 '18 at 18:37
30
$\begingroup$

Hint: The real line is the locus of points that are equidistant from $+i$ and $-i$.

enter image description here

$\endgroup$
43
$\begingroup$

Let $z = \frac{Z_1}{Z_2}\,$, then $|z-i|$ is the distance between point $z$ and point $i$ on the imaginary axis. Likewise, $|z+i|=|z-(-i)|$ is the distance between $z$ and $-i\,$. Therefore, $z$ is at equal distances from $i$ and $-i$, so it lies on the perpendicular bisector of the segment between $i$ and $-i$, which is in fact the real axis.

Alternatively, this can be proved algebraically as follows:

$$ |z-i|^2=|z+i|^2 \iff (z-i)(\bar z +i)=(z+i)(\bar z -i) \\ \iff |z|^2+1 +i(z-\bar z) = |z|^2+1+i(\bar z - z) \iff z-\bar z = 0 $$

But $z-\bar z = 2 i \operatorname{Im}(z)\,$, so $z-\bar z = 0$ means that $\operatorname{Im}(z)=0\,$ i.e. $\,z$ is a real number.

$\endgroup$
  • $\begingroup$ Ohhhhh right !! Thanks a lot !! $\endgroup$ – Aditi Jan 16 '18 at 17:46
  • $\begingroup$ Nice proofs. I suspect you could make the first one "algebraic" by letting $ \left| z - i \right| = k$ for some real $k$ and so on. $\endgroup$ – Carl Witthoft Jan 16 '18 at 19:29
  • 1
    $\begingroup$ @CarlWitthoft Thanks. The first one was meant for the geometric intuition, and I left the algebra for the second one. $\endgroup$ – dxiv Jan 16 '18 at 22:02
15
$\begingroup$

Even if you cannot think of a good shortcut (as given in the other answers), you can at least use the brute force: what's given is:

$$\frac{z_1-iz_2}{z_1+iz_2}\cdot\frac{\overline{z_1}+i\overline{z_2}}{\overline{z_1}-i\overline{z_2}}=1$$

or, after tidying up:

$$z_1\overline{z_1}+iz_1\overline{z_2}-i\overline{z_1}z_2+z_2\overline{z_2}=z_1\overline{z_1}-iz_1\overline{z_2}+i\overline{z_1}z_2+z_2\overline{z_2}$$

or, after cancelling identical terms and dividing by $2i$:

$$z_1\overline{z_2}=\overline{z_1}z_2$$

or, after dividing by $z_2\overline{z_2}$:

$$\frac{z_1}{z_2}=\overline{\left(\frac{z_1}{z_2}\right)}$$

i.e. $\frac{z_1}{z_2}$ is real.

Throughout I've used well-known properties of complex conjugate: $\overline{z_1\pm z_2}=\overline{z_1}\pm\overline{z_2}$, $\overline{z_1z_2}=\overline{z_1}\overline{z_2}$, $\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}$, $\overline{(\pm i)}=\mp i$, as well as the fact that $z$ is real if and only if $z=\overline{z}$.

$\endgroup$
  • $\begingroup$ Yes that’s also a good way to deal with it . Thank you ! $\endgroup$ – Aditi Jan 16 '18 at 17:54
10
$\begingroup$

Here's a geometric proof. The point $$B = \frac{Z_1 - iZ_2}{Z_1 - iZ_2}$$ is the image of the point $A = Z_1/Z_2$ under the Möbius transformation $$\mu(z) = \frac{z - i}{z + i}.$$ That means $A$ is the image of $B$ under the inverse transformation $\mu^{-1}$. You want to show that if $B$ lies on the unit circle, then $A$ lies on the real line. In other words, you want to show that $\mu^{-1}$ sends the unit circle to the real line. This is the same as showing that $\mu$ sends the real line to the unit circle.†

A Möbius transformation sends every circle, and every line, to either a circle or a line. To save ink, let's refer to both circles and lines as generalized circles, on the principle that a line is a "circle of infinite radius." Then we can just say a Möbius transformation sends every generalized circle to another generalized circle.

You can describe a generalized circle completely by listing three different points it passes through. The real line, for example, is the only generalized circle that passes through $-1$, $0$, and $1$. The unit circle is the only generalized circle that passes through $i$, $-1$, and $-i$.

You can show by direct calculation that $\mu$ sends the points $-1$, $0$, and $1$ to the points $i$, $-1$, and $i$, respectively. Since a Möbius transformation sends generalized circles to generalized circles, $\mu$ must send the generalized circle that passes through through $-1$, $0$, and $1$ to the generalized circle that passes through $i$, $-1$, and $-i$. In other words, $\mu$ must send the real line to the unit circle.


† To make this argument airtight, you have to see Möbius transformations as transformations of the extended complex plane: the complex plane plus an extra point $\infty$, which you can think of as the reciprocal of $0$.

$\endgroup$
  • $\begingroup$ +1 because complex analysis is awesome, but most "novices" might not think of this as being very "geometric". $\endgroup$ – einpoklum - reinstate Monica Jan 17 '18 at 20:56
9
$\begingroup$

Or (the "simpleton approach") you can let $\frac{Z_1}{Z_2} = x + yi$ where $x, y \in \mathbb R$

So you end up with $|x+(y+1)i| = |x+(y-1)i|$

which gives $x^2 + (y+1)^2 = x^2 + (y-1)^2 \implies (y+1)^2 - (y-1)^2 = 0 \implies 2y(2) = 0 \implies y = 0$. Since its imaginary part is zero, $\frac{Z_1}{Z_2}$ is real.

$\endgroup$

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.