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

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


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

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.

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    $\begingroup$ 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, 2018 at 17:46
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    $\begingroup$ The statement is not true for $Z_2=0$ (in that case, $Z_1/Z_2$ is not even defined) $\endgroup$
    – chtz
    Jan 17, 2018 at 9:23
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    $\begingroup$ @chtz: If $Z_2 = 0$, then the problem is ill-posed. $\endgroup$
    – Kevin
    Jan 18, 2018 at 4:23
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    $\begingroup$ @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, 2018 at 18:37

5 Answers 5


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.

  • $\begingroup$ Ohhhhh right !! Thanks a lot !! $\endgroup$
    – Aditi
    Jan 16, 2018 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. $\endgroup$ Jan 16, 2018 at 19:29
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    $\begingroup$ @CarlWitthoft Thanks. The first one was meant for the geometric intuition, and I left the algebra for the second one. $\endgroup$
    – dxiv
    Jan 16, 2018 at 22:02

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

enter image description here


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:


or, after tidying up:


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


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


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 $z=\overline{z}$.

  • $\begingroup$ Yes that’s also a good way to deal with it . Thank you ! $\endgroup$
    – Aditi
    Jan 16, 2018 at 17:54

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 $0$.

  • $\begingroup$ +1 because complex analysis is awesome, but most "novices" might not think of this as being very "geometric". $\endgroup$
    – einpoklum
    Jan 17, 2018 at 20:56

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.


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