Question

If $\sqrt{5-12i}+\sqrt{-5-12i}=z$, then principal value of $arg\left(z\right)$ can be

• A. $\frac{\pi }{4}$
• B. $\frac{3\pi }{4}$
• C. $-\frac{3\pi }{4}$
• D. $-\frac{\pi }{4}$

Answer 1

Answer: (A), (B), (C), (D)

$-\frac{\pi }{4}$

Given: $\sqrt{5-12i}+\sqrt{-5-12i}=z$

Square root of a complex number $a+ib$ is given by

$\sqrt{a+ib}$

$=\pm (\sqrt{\frac{1}{2}[\sqrt{a^2+b^2}+a]}-i\sqrt{\frac{1}{2}[\sqrt{a^2+b^2}}-a])$

$\Rightarrow \sqrt{5-12i}=\pm (3-2i)$ and

$\sqrt{-5-12i}=\pm (2-3i)$

$\Rightarrow z=-1-i,-5+5i,5-5i,1+i$

Therefore, principal values of $argz$ are $-\frac{3\pi }{4},\frac{3\pi }{4},-\frac{\pi }{4},\frac{\pi }{4}$