# Calculate square root of $i \Leftrightarrow z^2=i$

Let $z = r(\cos\theta+i\sin\theta)$. In my notes there was this example to calculate the square roots of $i$. What was done was:

$z = r(\cos\theta+i\sin\theta)\\z^2 = r^2(\cos(2\theta)+i\sin(2\theta))\\=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$

Im not entirely understand what they did above, what does the 2 lines below actually tell us? $\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$

I thought to calculate the square roots of iyou let $z=x+iy$ and work out $x,y$ from $(x+iy)^2=i$?

Also, how did they get that $r^2(\cos(2\theta)+i\sin(2\theta))=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})$?

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How did they get that $r^2(\cos(2\theta)+i\sin(2\theta))=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})$?

This whole problem is easiest to do via Euler's identity: $$e^{i\theta} = \cos(\theta) + i\sin(\theta).$$ Now, any complex number $z$ can be expressed as $z=re^{i\theta}$ where $r$ is some real number and $0 \leq \theta < 2\pi$. It follow that $z^2 = r^2e^{i2\theta} = r^2(\cos(2\theta)+i\sin(2\theta))$, and so if we want to solve the equation $z^2 = i$, then we have $$z^2 = r^2e^{i2\theta} = r^2(\cos(2\theta)+i\sin(2\theta)) = i.$$ Now, $r^2\cos(2\theta)+ir^2\sin(2\theta) = 0 + i$ means that $r^2\cos(2\theta) = 0$ and $r^2\sin(2\theta) = 1$. The latter equation shows us that $r \neq 0$ and hence it must be that $\cos(2\theta) = 0$, giving $2\theta = \pi/2$ and so $\sin(2\theta) = 1$. Substituting $\sin(2\theta) = 1$ into the latter equation reduces it to $r^2 = 1$, and so $r = \pm 1$. Also, since $2\theta = \pi/2$, we get $\theta = \pi/4$. Hence, the two square roots of $i$ are $$i^{1/2} = \pm\left(\cos(\pi/4) + i\sin(\pi(4)\right) = \pm \frac{1+i}{\sqrt{2}}$$

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Thanks for your help dilip, appreciated! –  Derrick May 23 '12 at 17:07
I understood how you got that $\cos(2\theta) = 0, \sin(2\theta) = 1$, but how did you deduce that $r = \pm 1$? –  Derrick May 23 '12 at 17:09
@Derrick See my revised answer. –  Dilip Sarwate May 23 '12 at 17:30
Thanks @dilipsarwate, I understand now, most appreciated again. :) –  Derrick May 23 '12 at 17:37

You can do it from $(x+iy)^2=i$, as follows: $$\displaylines{x^2+2ixy-y^2=i\cr x^2-y^2=0,\qquad2xy=1\cr y=\pm x,\qquad2x^2=\pm1\cr }$$ but $2x^2=-1$ is impossible for $x$ real, so $y=x$ and $2x^2=1$, yielding two solutions, $x+iy=\pm(\sqrt2/2+i\sqrt2/2)$.

It's good to know the other way, as it will work better for some other problems. In general, the more ways you know to do a problem, the better.

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Thanks for your advise, I wasnt aware of the other method until I saw it in my notes. –  Derrick May 23 '12 at 17:07

$i=1(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))$ and we want $z^2=i$, so $r^2(\cos(2\theta)+i\sin(2\theta))=1(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))$. Then use that $r_1(\cos\theta_1+i\sin\theta_1)=r_2(\cos\theta_2+i\sin\theta_2)\implies r_1=r_2\wedge\theta_1=\theta_2+2\pi k,k\in\mathbb{Z}$.

This gives $r^2=1\wedge2\theta=\frac{\pi}{2}+2\pi k,k\in\mathbb{Z}$. Since $r$ is a nonnegative real number, $r=1$; dividing both sides of $2\theta=\frac{\pi}{2}+2\pi k$ by 2 gives $\theta=\frac{\pi}{4}+\pi k$; picking values of $k$ starting from $k=0$ and stopping when we've gone full around gives $k=0,1$ (also, we should expect exactly 2 numbers whose square is $i$).

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Thanks for your explanation, appreciated! –  Derrick May 23 '12 at 17:06