What Is the Square Root of i in Complex Numbers?

[John O' Meara]

According to this equation \exp^{iy} = \cos y + \iota \sin y \mbox{ => } \exp^{\frac{\pi \iota}{2}} = \iota \ \mbox{,} \ \\ \exp^{\pi \iota} = -1 \ \ \exp^{\frac{-\pi \iota}{2}} = -\iota \ \ \exp^{2 \pi \iota} = 1 \\. What then is \sqrt{\iota} = \iota^{\frac{1}{2}} \mbox{ equal to ?} \\
Does not \exp^{\frac{\pi \iota}{4}}=\cos\frac{\pi}{4} + \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} +\frac{\iota}{\sqrt{2}} \ \mbox{ which is not =} \ \sqrt{\iota} \\
I find it hard to see that it equals \exp^{\frac{\pi \iota}{4}}\\. Help would be welcome. Thanks.

 

[Dick]

(1/sqrt(2)+i/sqrt(2))^2=1/2+2*i/(sqrt(2)*sqrt(2))-1/2=2*i/2=i.

 

[HallsofIvy]

According to this equation \exp^{iy} = \cos y + \iota \sin y \mbox{ => } \exp^{\frac{\pi \iota}{2}} = \iota \ \mbox{,} \ \\ \exp^{\pi \iota} = -1 \ \ \exp^{\frac{-\pi \iota}{2}} = -\iota \ \ \exp^{2 \pi \iota} = 1 \\. What then is \sqrt{\iota} = \iota^{\frac{1}{2}} \mbox{ equal to ?} \\
Does not \exp^{\frac{\pi \iota}{4}}=\cos\frac{\pi}{4} + \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} +\frac{\iota}{\sqrt{2}} \ \mbox{ which is not =} \ \sqrt{\iota} \\
I find it hard to see that it equals \exp^{\frac{\pi \iota}{4}}\\. Help would be welcome. Thanks.

Why do you think \sqrt{2}/2 +i\sqrt{2}/2 is not equal to \sqrt{i}. Dick did the calculations but it's just a little bit hard to read. In Tex:
\left(\frac{\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\right)
= \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}\frac{i\sqrt{2}}{2}+ \frac{i\sqrt{2}}{2}\frac{i\sqrt{2}}{2}
= \frac{1}{2}+ \frac{i}{2}+ \frac{i}{2}- \frac{1}{2}= i
Of course, like any non-zero number, i has two square roots. The other is
-\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}.