Question

If $\left|\frac{z_1-i{z}_2}{z_1+i{z}_2}\right|=1,\text{then}\frac{z_1}{z_2}\text{is}$

• A. purely imaginary
• B. real
• C. imaginary
• D. of unit modulus

Answer 1

The correct option is B real

Given,

$\left|\frac{z_1-i{z}_2}{z_1+i{z}_2}\right|=1$

Now squaring both sides we get,

$|z_1-i{z}_2{|}^2=|z_1+i{z}_2{|}^2$

or, $(z_1-i{z}_2)(\overline{z_1}+i\overline{z_2})$$=(z_1+i{z}_2)(\overline{z_1}-i\overline{z_2})$ [ Since $|z{|}^2=z\overline{z}$]

or, $|z_1{|}^2+|z_2{|}^2+i(z_1\overline{z_2}-z_2\overline{z_1})$$=|z_1{|}^2+|z_2{|}^2-i(z_1\overline{z_2}-z_2\overline{z_1})$

or, $2i(z_1\overline{z_2}-z_2\overline{z_1})=0$

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

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

This gives $\frac{z_1}{z_2}$ is purely real.