Question

If $z_1,z_2$ are the complex numbers such that $|z_1+z_2|=|z_1|+|z_2|$ then $arg{z}_1-arg{z}_2$ is

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

Answer 1

The correct option is C $0$

We have,

$|z_1+z_2|=|z_1|+|z_2|$

Then, $\arg \left(z_1\right)-\arg \left(z_2\right)$

Let$z_1=\cos {\theta }_1+i\sin {\theta }_1$

$z_2=\cos {\theta }_2+i\sin {\theta }_2$

$\therefore |z_1+z_2|=|z_1|+|z_2|$

$\Rightarrow \sqrt{{(cos\theta +\cos {\theta }_2)}^2+{(sin{\theta }_1+\sin {\theta }_2)}^2}=|z_1|+|z_2|$

$|z_1|=\sqrt{{\cos }^2{\theta }_1+\sin {\theta }_1}=1$

$|z_2|=\sqrt{{\cos }^2{\theta }_2+\sin {\theta }_2}=1$

Now, on squaring and we get.

$\Rightarrow 2[1+\cos ({\theta }_1+{\theta }_2\left)\right]={(1+1)}^2$\

$\Rightarrow 2+2\cos ({\theta }_1-{\theta }_2)=4$

$\Rightarrow 2\cos ({\theta }_1-{\theta }_2)=2$

$\Rightarrow \cos ({\theta }_1-{\theta }_2)=1$

$\Rightarrow \cos ({\theta }_1-{\theta }_2)=\cos 0^0$

$\Rightarrow {\theta }_1-{\theta }_2=0^0$

$\arg z_1-\arg z_2=0$

Hence, this is the answer.