Solving $z^2 =\sqrt{\sqrt{i}}$

Question

I need to find a number $z \in \mathbb{C}$ such that $z^{2}=\sqrt{\sqrt{i}}$.

Let $z^{2}=(x+iy)^2=(x^2-y^2)+i(2xy)=\sqrt{i}$. I already know that $\sqrt{i}=\pm\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)$. Then $x^{2}-y^{2}=\frac{1}{\sqrt{2}}$ and $2xy=\frac{1}{\sqrt{2}}$, but I'm stuck finding the values of $x$ and $y$.

Can anyone help me finding the right values of $\sqrt{\sqrt{i}}$? Thanks

Answer 1

Polar form is much easier.

$i$ has $r = 1$ and $\theta = \frac{\pi}{2}$. Since when you square a number, you square the radius and double the angle, in this case you square root the radius and halve the angle.

Taking the square root gives $r = 1$ and $\theta = \frac{\pi}{4}$, and doing this two more times gives $r = 1$ and $\theta = \frac{\pi}{16}$. Converting back to Cartesian, (from Wolfram Alpha), $x = 1 \cdot \cos \left(\frac{\pi}{8} \right) = \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}$, and $y = 1 \cdot \sin \left(\frac{\pi}{8} \right) = \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}}$, so one root is $\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}} + \frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2}}}i$.

There are other roots as well. However, this is just one of the roots.

Answer 2

You need to solve two systems of equations: $$1) \ \begin{cases} x^{2}-y^{2}=\frac{1}{\sqrt{2}}\\ 2xy=\frac{1}{\sqrt{2}}\end{cases} \ \ \ \text{and} \ \ \ 2) \ \begin{cases} x^{2}-y^{2}=-\frac{1}{\sqrt{2}}\\ 2xy=-\frac{1}{\sqrt{2}}\end{cases}.$$ Case 1: $$y=\frac{1}{2\sqrt{2}x} \Rightarrow x^2-\frac{1}{8x^2}=\frac{1}{\sqrt{2}} \Rightarrow 8x^4-4\sqrt{2}x^2-1=0 \Rightarrow \\ x^2=\frac{\sqrt{2}+2}{4} \Rightarrow x_{1,2}=\pm\frac{\sqrt{2+\sqrt{2}}}{2} \Rightarrow y_{1,2}=\pm\frac{\sqrt{2-\sqrt{2}}}{2}.$$ Similarly, you can do Case 2.

Answer 3

Polar coordinates make this simple.

$\arg i = \frac \pi 2$ so $i = e^{\frac \pi2i+2k\pi}= e^{\frac \pi 2i}$ When viewed as angles $\frac \pi 2= \frac \pi 2 + 2k\pi$ so $\frac\pi 2$ is the only one we need to take note of.

So $\sqrt i = e^{i\theta}$ where $2\theta = \frac \pi 2+2k \pi$. So $\theta = \frac \pi 4 + k\pi$. $k$ can be even or odd. When viewed as angles if $k$ is even the $\frac \pi 4 + k\pi =\frac \pi 4$. And if $k$ is odd then $\frac \pi 4 + k\pi = \frac \pi 4 + \pi$ so $\theta =\frac \pi 4$ or $\frac \pi 4 + \pi$ are the only two we need to take note of.

So $\sqrt i = e^{\frac \pi 4i}$ or $e^{\frac \pi 4i + \pi i}$. (Note: $e^{\frac \pi 4i} = \frac {\sqrt 2}2 + \frac {\sqrt 2}2i$ and $e^{\frac \pi 4i + \pi i}= -\frac {\sqrt 2}2 - \frac {\sqrt 2}2i$ which is what we expected if we had used the rectangular coordinates of the $a + bi$ form. Polar coordinaties make multiplication, and division and roots and powers much easier (but addition and subtraction much harder).

$\sqrt{\sqrt i} = e^{\theta i}$ where $2\theta=\frac \pi 4 + k\pi$ and That is $\theta = \frac \pi 8 + \frac k2\pi$. As angles, if $k=0,1,2,3$ then $\frac \pi 8 + \frac k2\pi$ are different from each other. But as $\frac {k+4}2\pi = \frac k2\pi + 2\pi=\frac k2\pi$ the values repeat and $\sqrt{\sqrt i} = e^{(\frac k8\pi + \frac k2\pi)i}$ are four values.

At this point I think it is safe to say $\sqrt{\sqrt i} =\sqrt[4] i$ and there are $4$ values.

And that if $z^2 =\sqrt[4]{i}$ then $z = \sqrt[8]{i}$ and there will be $8$ values. And as $i =e^{(\frac \pi 2 + 2k\pi)i}$ then $\sqrt[8]{i} = e^{\frac {\frac \pi 2 + 2k\pi}2 i} = e^{(\frac \pi{16} + \frac k4\pi)i}$. And for $k=0.....,7$ these will be different angles and those are the eight roots.

If you want to put them back to rectangular coordinates.

$\sqrt[8]{i} = e^{(\frac \pi{16} + \frac k4\pi)i} = \cos(\frac \pi{16} + \frac k4\pi) + i \sin(\frac \pi{16} + \frac k4\pi)$ for $k= 0.....7$.