Solve this

Question:

If $z^{2}+|z|^{2}=0$, show that $z$ is purely imaginary.

 

Solution:

Let $z=a+i b$

$\left.\Rightarrow|z|=\sqrt{(} a^{2}+b^{2}\right)$

Now, $z^{2}+|z|^{2}=0$

$\Rightarrow(a+i b)^{2}+a^{2}+b^{2}=0$

$\Rightarrow a^{2}+2 a b i+i^{2} b^{2}+a^{2}+b^{2}=0$

$\Rightarrow a^{2}+2 a b i-b^{2}+a^{2}+b^{2}=0$

$\Rightarrow 2 a^{2}+2 a b i=0$

$\Rightarrow 2 a(a+i b)=0$

Either a = 0 or z = 0

Since $z \neq 0$

$a=0 \Rightarrow z$ is purely imaginary.

 

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