Question

If $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$, then find $\arg \left(z_1\right)-\arg \left(z_2\right)$.

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Step 1. Assume two complex numbers.

Let $z_1=\cos {\theta }_1+i\sin {\theta }_1$ and $z_2=\cos {\theta }_2+i\sin {\theta }_2$ be two complex numbers.

The argument of $z_1$ is $\arg \left(z_1\right)={\theta }_1$, and the argument of $z_2$ is $\arg \left(z_2\right)={\theta }_2$.

The modulus of $z_1$ and $z_2$ is $\left|z_1\right|=\left|z_2\right|=1$.

Step 2. Find the value of $\left|z_1+z_2\right|$.

In the expression $\left|z_1+z_2\right|$ substitute $z_1=\cos {\theta }_1+i\sin {\theta }_1$ and $z_2=\cos {\theta }_2+i\sin {\theta }_2$.

$\begin{array}{rcl}\left|z_1+z_2\right|& =& \left|\cos {\theta }_1+i\sin {\theta }_1+\cos {\theta }_2+i\sin {\theta }_2\right|\\ & =& \left|\cos {\theta }_1+\cos {\theta }_2+i\left(\sin {\theta }_1+\sin {\theta }_2\right)\right|\\ & =& \sqrt{{\left(\cos {\theta }_1+\cos {\theta }_2\right)}^2+{\left(\sin {\theta }_1+\sin {\theta }_2\right)}^2}\\ & =& \sqrt{{\cos }^2{\theta }_1+{\cos }^2{\theta }_2+2\cos {\theta }_1\cos {\theta }_2+{\sin }^2{\theta }_1+{\sin }^2{\theta }_2+2\sin {\theta }_1\sin {\theta }_2}\\ & =& \sqrt{\left({\cos }^2{\theta }_1+{\sin }^2{\theta }_1\right)+\left({\cos }^2{\theta }_2+{\sin }^2{\theta }_2\right)+2\left(\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2\right)}\\ & =& \sqrt{1+1+2\cos \left({\theta }_1-{\theta }_2\right)}\left[{\cos }^{2}A+{\sin }^{2}A=1,\cos A\cos B+\sin A\sin B=\cos \left(A-B\right)\right]\\ & =& \sqrt{2+2\cos \left({\theta }_1-{\theta }_2\right)}\end{array}$

Step 3 Find the value of $\arg \left(z_1\right)-\arg \left(z_2\right)$.

In the equation $\left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|$, substitute the values of $\left|z_1+z_2\right|$, $\left|z_1\right|$ and $\left|z_2\right|$.

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

Now square both sides.

$$\begin{gathered} {\left(\sqrt{2+2\cos \left({\theta }_1-{\theta }_2\right)}\right)}^2=2^2 \\ \Rightarrow 2+2\cos \left({\theta }_1-{\theta }_2\right)=4 \\ \Rightarrow 2\cos \left({\theta }_1-{\theta }_2\right)=2 \\ \Rightarrow \cos \left({\theta }_1-{\theta }_2\right)=1 \\ \Rightarrow {\theta }_1-{\theta }_2=0\left[\cos 0=1\right] \end{gathered}$$

Now, since $\arg \left(z_1\right)={\theta }_1$ and $\arg \left(z_2\right)={\theta }_2$ and it is found that $\begin{array}{rcl}{\theta }_1-{\theta }_2& =& 0\end{array}$.

So, $\arg \left(z_1\right)-\arg \left(z_2\right)=0$

Thus the value of $\arg \left(z_1\right)-\arg \left(z_2\right)$ is $0$.

Hence, the correct option is A.