Proving that a complex number is real

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.

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

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 $z=\overline{z}$.