MCQ
$|z_1 + z_2| = |z_1| + |z_2|$ is possible if:
  • A
    $\text{z}_2=\bar{\text{z}_1}$
  • B
    $\text{z}_2=\frac{1}{\text{z}_1}$
  • $\arg(\text{z}_1)=\arg(\text{z}_2)$
  • D
    $|\text{z}_1|=|\text{z}_2|$

Answer

Correct option: C.
$\arg(\text{z}_1)=\arg(\text{z}_2)$
$\text{arg}(\text{z}_1)=\text{arg}(\text{z}_2)$
Let $\text{z}_1=\text{r}_1(\cos\theta_1+\text{i}\sin\theta_1)$ and $\text{z}_2=\text{r}_2(\cos\theta_2+\text{i}\sin\theta_2)$
Since $|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
$|\text{z}_1+\text{z}_2|=\text{r}_1\cos\theta_1+\text{i}\text{r}_1\sin\theta_1+\text{r}_2\cos\theta_2+\text{i}\text{r}_2\sin\theta_2$
$|\text{z}_1+\text{z}_2|=\sqrt{\text{r}^2_1\cos^2\theta_1+\text{r}^2_2\cos^2\theta_2+2\text{r}_1\text{r}_2\cos\theta_1\cos\theta_2\\+\text{r}^2_1\sin^2\theta_1+\text{r}^2_2\sin^2\theta_2+2\text{r}_1\text{r}_2\sin\theta_1\sin\theta_2}$
$=\sqrt{\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)}$
But $|\text{z}_1+\text{z}_2|=|\text{z}_1|+|\text{z}_2|$
So, $\sqrt{\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)}=\text{r}_1+\text{r}_2$
Squaring both sides, we get
$\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)=\text{r}^2_1+\text{r}^2_2+2\text{r}_1\text{r}_2$
$\Rightarrow2\text{r}_1\text{r}_2-2\text{r}_1\text{r}_2\cos(\theta_1-\theta_2)=1$
$\Rightarrow1-\cos(\theta_1-\theta_2)=0$
$\Rightarrow\cos(\theta_1-\theta_2)=1$
$\Rightarrow\theta_1-\theta_2=0$
$\Rightarrow\theta_1=\theta_2$
So, $\text{arg}(\text{z}_1)=\text{arg}(\text{z}_2)$

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