Question

Taking the value of a square root with positive real part only, the value of $\sqrt{-3-4i}+\sqrt{3+4i}$ is

• A. $1+i$
• B. $1-3i$
• C. $1+3i$
• D. none of these

Answer 1

The correct option is D none of these

We need to find value of $\sqrt{-3-4i}+\sqrt{3+4i}$.

$(r{e}^{i(\theta -\pi )}{)}^{\frac{1}{2}}+(r{e}^{i\theta }{)}^{\frac{1}{2}}$
$=\sqrt{r}[e^{i\frac{\theta -\pi }{2}}+e^{i\frac{\theta }{2}}]$
$=\sqrt{5}\left[\cos \right(\frac{\theta -\pi }{2})+i\sin (\frac{\theta -\pi }{2})+\cos (\frac{\theta }{2})+i\sin (\frac{\theta }{2}\left)\right]$
$=\sqrt{5}\left[\cos \right(\frac{\theta }{2})+\sin (\frac{\theta }{2})+i(\sin \left(\frac{\theta }{2}\right)-\cos \left(\frac{\theta }{2}\right)\left)\right]$
Now $\theta ={53}^0$
Hence, the above expression reduces to $3-i$.
Hence, the answer is none of the above.