# Very simple complex question

• John O' Meara

#### 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.

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

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
$$-\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$$.