# $$\sqrt{i}$$

Why is $$\sqrt{i} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$$

5 years, 2 months ago

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$$\displaystyle i=e^{i\pi/2} \Rightarrow \sqrt{i}=e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$

- 5 years, 2 months ago

remember taking root will give both positive and negative solutions

- 5 years, 2 months ago

I completely agree though you forgot to put in the $$i$$ until the end: $$i=e^{\pi i/2}$$ Voted up. :D

- 5 years, 2 months ago

Thanks! :)

- 5 years, 2 months ago

Actually, $$i$$ has two square roots, just as any real number does. These are the two roots of $$z^2=e^{\pi i/2}$$. One of the roots is $$e^{\pi i/4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$ as noted previously. The other is $$e^{5\pi i/4}=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$$. Doubling the argument of both of these complex numbers and squaring the modulus indeed gives $$e^{\pi i/2}$$.

- 5 years, 2 months ago

You're missing a minus sign:

$$e^{5 \pi i / 4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$$

- 5 years, 2 months ago

There is some method with which we agree on which root we're talking about with radicals, making radicals ambiguous but fractional exponents not so.

Sadly, I don't know exactly how we do decide. Could anyone enlighten me?

- 5 years, 2 months ago

The "principal root" of x describes the positive root, so if you're given $$\sqrt{4}$$ then it is asking for the positive answer which is $$2$$ and not $$-2$$. but when dealing with complex numbers it gets.. well, more complex. Neither root is more "important" than the other, so you don't tend to worry about it as much.

- 5 years, 2 months ago

and for higher roots? (e.g. $$\sqrt[5]{2}$$)

- 5 years, 2 months ago

let $$\sqrt{i}$$ = $$x$$

we have $$x^{2}$$ = $$i$$

converting i to polar form, we have

$$x^{2}$$ = $$\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}$$

using Demoivre's Theorem

we have $$x_{1} = \cos \frac{\pi}{4} + i\sin \frac{\pi}{4} = \frac{\sqrt2}{2} + \frac{\sqrt2}{2} i$$ and $$x_{2} = \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} = -\frac{\sqrt2}{2} -\frac{\sqrt2}{2} i$$

- 5 years, 2 months ago

$$\sqrt{i}$$ is a complex number, so it can be written in the form $$a+bi$$ where $$a$$ and $$b$$ are real numbers.

Squaring gives $$(a^2-b^2)+2abi$$. It follows that $$2ab=1$$ and $$a^2=b^2$$.

From the second we have that $$a=b$$ or $$a=-b$$.

Substituting in the first, we get $$a^2=\frac{1}{2}$$ or $$a^2=-\frac{1}{2}$$

So $$a=\pm \frac{1}{2} \sqrt{2}$$ or $$a=\pm \frac{1}{2} \sqrt{2}*i$$.

and $$b=\pm \frac{1}{2} \sqrt{2}$$ or $$b=\mp \frac{1}{2} \sqrt{2}*i$$.

Using these in our original expression $$a+bi$$, we get only two distinct solutions:

$$\frac{1}{2} \sqrt{2} + \frac{1}{2} \sqrt{2}*i$$ and $$- \frac{1}{2} \sqrt{2} - \frac{1}{2} \sqrt{2}*i$$.

I think usually the first is chosen as THE root of $$i$$, but only because it's closest to $$(1,0)$$.

- 5 years, 2 months ago