Imaginary Unit

Converting to polar coordinates, we have that the modulus of $i$ is: $|i| = \sqrt{1^2} = 1$ . So $i = 0 + i = \cos\left(\frac{\pi}{2} + 2\pi k\right) + i \sin\left(\frac{\pi}{2} + 2\pi k\right) = e^{i\cdot \left( \frac{\pi}{2} + 2\pi k \right)}$ , for any integer $k$ . Thus

$\sqrt{i} = \sqrt{e^{i\cdot \left( \frac{\pi}{2} + 2\pi k\right)}} = e^{i\cdot \left(\frac{\pi}{4}+\pi k\right)} = \cos\left(\frac{\pi}{4} + \pi k\right) + i \sin\left(\frac{\pi}{4} + \pi k\right)$

Since $0 \leq \theta \leq \frac{\pi}{2}$ , thus $\theta = \frac{\pi}{4}$ . Converting to degrees, we have $\theta = \frac{\pi}{4} \cdot \frac{180^\circ}{\pi} = 45 ^\circ$ .

$\sqrt{i}$ =r( $\cos\theta$ +i $\sin\theta$ )

Square both side of equation:

i= $r^{2}$ ( $\cos^{2}\theta$ - $\sin^{2}\theta$ +2 $\sin\theta \cos\theta$ )

= $r^{2}$ $\cos 2\theta$ + $r^{2}$ i $\sin 2\theta$

The real part is zero, therefore $\cos 2\theta$ =0, $\theta$ =45 degrees