Question

If $z_1$ and $z_2$ are two non-zero complex numbers such that $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$, then $\arg \left(\frac{z_1}{z_2}\right)$ is equal to

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

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Find the argument of $\frac{z_1}{z_2}$.

Let $z_1=\cos {\theta }_1+i\sin {\theta }_1$ and $z_2=\cos {\theta }_2+i\sin {\theta }_2$ be two complex numbers. Now, $\arg \left(z_1\right)={\theta }_1$ and $\arg \left(z_2\right)={\theta }_2$.

The magnitude of $z_1$ is

$\begin{array}{rcl}\left|z_1\right|& =& \left|\cos {\theta }_1+i\sin {\theta }_1\right|\\ & =& \sqrt{{\cos }^2{\theta }_1+{\sin }^2{\theta }_1}\\ & =& \sqrt{1}\\ & =& 1\end{array}$

The magnitude of $z_2$ is

$\begin{array}{rcl}\left|z_2\right|& =& \left|\cos {\theta }_2+i\sin {\theta }_2\right|\\ & =& \sqrt{{\cos }^2{\theta }_2+{\sin }^2{\theta }_2}\\ & =& \sqrt{1}\\ & =& 1\end{array}$

The complex number $z_1+z_2$ is given as:

$\begin{array}{rcl}{z}_1+z_2& =& \cos {\theta }_1+i\sin {\theta }_1+\cos {\theta }_2+i\sin {\theta }_2\\ & =& \left(\cos {\theta }_1+\cos {\theta }_2\right)+i\left(\sin {\theta }_1+\sin {\theta }_2\right)\end{array}$

The magnitude of $z_1+z_2$ is given as:

$\begin{array}{rcl}\left|z_1+z_2\right|& =& \left|\left(\cos {\theta }_1+\cos {\theta }_2\right)+i\left(\sin {\theta }_1+\sin {\theta }_2\right)\right|\\ & =& \sqrt{{\left(\cos {\theta }_1+\cos {\theta }_2\right)}^2+{\left(\sin {\theta }_1+\sin {\theta }_2\right)}^2}\\ & =& \sqrt{{\cos }^2{\theta }_1+{\cos }^2{\theta }_2+2\cos {\theta }_1\cos {\theta }_2+{\sin }^2{\theta }_1+{\sin }^2{\theta }_2+2\sin {\theta }_1\sin {\theta }_2}\\ & =& \sqrt{\left({\cos }^2\theta {}_1+{\sin }^2\theta {}_1\right)+\left(\cos {}^2\theta {}_2+\sin {}^2\theta {}_2\right)+2\left(\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2\right)}\\ & =& \sqrt{1+1+2\cos \left({\theta }_1-{\theta }_2\right)}\left[{\cos }^{2}A+{\sin }^{2}A=1,\cos \left(A-B\right)=\cos A\cos B+\sin A\sin B\right]\\ & =& \sqrt{2+2\cos \left({\theta }_1-{\theta }_2\right)}\end{array}$Now, substitute the values of $\left|z_1\right|,\left|z_2\right|,\left|z_1+z_2\right|$ in the equation $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$ and simplify by squaring both sides.

$\begin{array}{rcl}\left|z_1+z_2\right|& =& \left|z_1\right|+\left|z_2\right|\\ & \Rightarrow & \begin{array}{rcl}& & \sqrt{2+2\cos \left({\theta }_1-{\theta }_2\right)}\end{array}=1+1\\ & \Rightarrow & {\left(\begin{array}{rcl}& & \sqrt{2+2\cos \left({\theta }_1-{\theta }_2\right)}\end{array}\right)}^2=2^2\\ & \Rightarrow & \begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \left({\theta }_1-{\theta }_2\right)\end{array}=\begin{array}{rcl}& & 4\end{array}\\ \begin{array}{rcl}& & \end{array}& \Rightarrow & \begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \left({\theta }_1-{\theta }_2\right)\end{array}=\begin{array}{rcl}& & 2\end{array}\\ \begin{array}{rcl}& & \end{array}& \Rightarrow & \begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \cos \end{array}\begin{array}{rcl}& & \left({\theta }_1-{\theta }_2\right)\end{array}=\begin{array}{rcl}& & 1\end{array}\begin{array}{rcl}& & \end{array}\\ \begin{array}{rcl}& & \end{array}& \Rightarrow & \begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \theta \end{array}\begin{array}{rcl}& & {}_1\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \theta \end{array}\begin{array}{rcl}& & {}_2\end{array}=\begin{array}{rcl}& & 0\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \left[\cos 0=1\right]\end{array}\end{array}$

Now, $\arg \left(\frac{z_1}{z_2}\right)$ is given as:

$\begin{array}{rcl}\arg \left(\frac{z_1}{z_2}\right)& =& \arg \left(z_1\right)-\arg \left(z_2\right)\\ & =& {\theta }_1-{\theta }_2\\ & =& 0\end{array}$

So, the value of $\arg \left(\frac{z_1}{z_2}\right)$ is $0$.

Hence, the correct option is A.