Question

Let $z_1$ and $z_2$ be complex number such that $z_1\ne z_2$ and $\left|z_1\right|=\left|z_2\right|.$ If $z_1$ has positive real part and $z_2$ has negative imaginary part, then $\frac{z_1+z_2}{z_1-z_2}$ may be

• A. $0$
• B. real and positive
• C. real and negative
• D. purely imaginary

Answer 1

The correct option is D purely imaginary

Let $z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$

where, $x_1\ne x_2,y_1\ne y_2$ and ${x_1}^2+{y_1}^2={x_2}^2+{y_2}^2$

Now, $\frac{z_1+z_2}{z_1-z_2}=\frac{(x_1+x_2)+i(y_1+y_2)}{(x_1-x_2)+i(y_1-y_2)}$

$=\frac{[(x_1+x_2)+i(y_1+y_2)][(x_1-x_2)-i(y_1-y_2)]}{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

$=\frac{[({x_1}^2-{x_2}^2)+({y_1}^2-{y_2}^2)]+i[x_1{y}_1-y_1{x}_2+y_2{x}_1-y_2{x}_2-x_1{y}_1+x_1{y}_2-x_2{y}_1+x_2{y}_2]}{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

$=\frac{2i(x_1{y}_2-y_1{x}_2)}{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

$=$ a purely imaginary or $0$ if $\frac{x_1}{x_2}=\frac{y_1}{y_2}$

If $\frac{x_1}{x_2}=\frac{y_1}{y_2}$ then $x_1+i{y}_1=k(x_2+i{y}_2)$

If $k=1,z_1=z_2$, which is not true and if $k\ne 1,\left|z_1\right|\ne \left|z_2\right|$

$\therefore \frac{z_1+z_2}{z_1-z_2}$ is purely imaginary.