Question

What is the value of $√-i$, where$i=√-1$​?

Answer 1

Answer:

$\displaystyle \sqrt{-i}=-\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2}$

Step-by-step explanation:

We want to find the value for $\sqrt{-i}$.

In the Real and Imaginary axis diagram, we can notice that the complex number i is 90 degrees anticlockwise from the positive Real axis. Therefore we can define i as:

$\displaystyle i=e^{i\frac{\pi}{2}}$

Substitute the value of i to $\sqrt{-i}$:

$\sqrt{-i}=\sqrt{-e^{i\frac{\pi}{2}}$

$\sqrt{-i}=\sqrt{-1}\cdot\sqrt{e^{i\frac{\pi}{2}}}$

$\sqrt{-i}=ie^{i\frac{\pi}{4}}$

Using Euler's formula:

$e^{i\theta}=\cos \theta+i\sin \theta$

$e^{i\frac{\pi}{4} }=\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}$

$e^{i\frac{\pi}{4} }=\frac{1}{2}\sqrt{2} +i\frac{1}{2}\sqrt{2}$

Hence:

$\sqrt{-i}=ie^{i\frac{\pi}{4}}$

$\displaystyle \sqrt{-i}=i\left(\frac{1}{2}\sqrt{2} +i\frac{1}{2}\sqrt{2}\right)$

$\displaystyle \sqrt{-i}=\left(i\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}\right)$

Therefore, the square root of -i is:

$\displaystyle \sqrt{-i}=-\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2}$