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If $z_{1}, z_{2}$ are any two complex numbers such that $\left|z_{1}+z_{2}\right|=$ $\left|z_{1}\right|+\left|z_{2}\right|$, which one of the following is correct ?
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Verified Answer
The correct answer is:
$\mathrm{z}_{1}=\alpha \mathrm{z}_{2}$ with $\alpha \in \mathrm{R}$
Let $z_{1}=r_{1}(\cos \theta+i \sin \theta)$
Squaring both the sides $\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}+2 \mathrm{r}_{1} \mathrm{r}_{2} \cos (\theta-\phi)=\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)^{2}$
$=\mathrm{r}_{1}{ }^{2}+\mathrm{r}_{2}{ }^{2}+2 \mathrm{r}_{1} \mathrm{r}_{2}$
or, $2 \mathrm{r}_{1} \mathrm{r}_{2} \cos (\theta-\phi)=2 \mathrm{r}_{1} \mathrm{r}_{2}$
or, $\cos (\theta-\phi)=1$
$\theta-\phi=0 \Rightarrow \theta=\phi$
$\operatorname{Arg}\left(z_{1}\right)=\operatorname{Arg}\left(z_{2}\right)$ so, $z_{1}=\alpha z_{2}$
where $\alpha \in \mathrm{R}$
Squaring both the sides $\mathrm{r}_{1}^{2}+\mathrm{r}_{2}^{2}+2 \mathrm{r}_{1} \mathrm{r}_{2} \cos (\theta-\phi)=\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)^{2}$
$=\mathrm{r}_{1}{ }^{2}+\mathrm{r}_{2}{ }^{2}+2 \mathrm{r}_{1} \mathrm{r}_{2}$
or, $2 \mathrm{r}_{1} \mathrm{r}_{2} \cos (\theta-\phi)=2 \mathrm{r}_{1} \mathrm{r}_{2}$
or, $\cos (\theta-\phi)=1$
$\theta-\phi=0 \Rightarrow \theta=\phi$
$\operatorname{Arg}\left(z_{1}\right)=\operatorname{Arg}\left(z_{2}\right)$ so, $z_{1}=\alpha z_{2}$
where $\alpha \in \mathrm{R}$
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