Question

If $z_1=\overline{z_2}$ and $z_3=\overline{z_4}$, then $arg\left(\frac{z_4}{z_1}\right)+arg\left(\frac{z_3}{z_2}\right)$ is equal to:

• A. $\pi$
• B. 0
• C. $\frac{3\pi }{2}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\pi$

Let $z_1=x_1+i{y}_1$, $z_2=x_2+i{y}_2$, $z_3=x_3+i{y}_3$, $z_4=x_4+i{y}_4$

Given that $z_1={\overline{z}}_2$ and $z_3={\overline{z}}_4$

$\Rightarrow x_1+i{y}_1=x_2-i{y}_2$ and $x_3+i{y}_3=x_4-i{y}_4$

Therefore, $x_1=x_2,y_1=-y_2,x_3=x_4,y_3=-y_4$

Now, $\Rightarrow \frac{z_4}{z_1}=\frac{x_4+i{y}_4}{x_1+i{y}_1}$

$\Rightarrow \frac{z_4}{z_1}=\frac{x_4+i{y}_4}{x_1+i{y}_1}\times \frac{x_1-i{y}_1}{x_1-i{y}_1}$

$\Rightarrow \frac{z_4}{z_1}=\frac{x_1{x}_4+y_1{y}_4+i(x_1{y}_4-x_4{y}_1)}{x_1^2+y_1^2}$

Similarly, $\Rightarrow \frac{z_3}{z_2}=\frac{x_2{x}_3+y_2{y}_3+i(x_2{y}_3-x_3{y}_2)}{x_2^2+y_2^2}$

But, $x_1=x_2,y_1=-y_2,x_3=x_4,y_3=-y_4$

Therefore, $\Rightarrow \frac{z_3}{z_2}=\frac{x_1{x}_4+y_1{y}_4+i(-x_1{y}_4+x_4{y}_1)}{x_1^2+y_1^2}$

$\Rightarrow arg\left(\frac{z_4}{z_1}\right)={\tan }^{-1}\left(\frac{x_1{y}_4-x_4{y}_1}{x_1{x}_4+y_1{y}_4}\right)$

$\Rightarrow arg\left(\frac{z_3}{z_2}\right)={\tan }^{-1}\left(\frac{-x_1{y}_4+x_4{y}_1}{x_1{x}_4+y_1{y}_4}\right)$

$\Rightarrow arg\left(\frac{z_3}{z_2}\right)={\tan }^{-1}\left(\frac{-(x_1{y}_4-x_4{y}_1)}{x_1{x}_4+y_1{y}_4}\right)$

We know that ${\tan }^{-1}x+{\tan }^{-1}(-x)=\pi$

Therefore, $\Rightarrow arg\left(\frac{z_4}{z_1}\right)+arg\left(\frac{z_3}{z_2}\right)=\pi$