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I need to simplify $z^2+i=0$ and find all solutions for $z$. I have seen that the solutions to $z=\sqrt{i}=\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)$. I was hoping to find a similar solution for $z=\sqrt{-i}\,$ but my attempt gives me $z=\pm i^{\frac{3}{2}}$

$$z=re^{i\theta} \,\,\& \,\, e^{i\pi}=-1 $$ then, $$(re^{i\theta})^2=-i\\r^2e^{i2\theta}=ie^{i\pi+k(2\pi)}$$ where $k\in\mathbb{Z}$.

So, we have $\begin{cases} r^2=i \,\,\,\therefore r=\sqrt{i}\\ \theta=k\pi \end{cases}$

Then, $$z_k=\sqrt{i} \, e^{i\left(\frac{\pi}{2}+k\pi\right)}$$

$$z_0=\sqrt{i}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=i^\frac{3}{2}\\ z_1=\sqrt{i}\left(\cos(\frac{\pi}{2}+1)+i(\sin\frac{\pi}{2}+1)\right)=-i^\frac{3}{2}$$

I realized that this is literally the same as just solving $z=\sqrt{-i}=i\sqrt{i}=i^\frac{3}{2}$, however, I was hoping to find a solution of the form $x+iy$. I am not sure how to go about this problem a different way.

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  • $\begingroup$ How did you find the solutions for $\sqrt{i}$ (or how were they derived)? What do you know about polar form of complex numbers? $\endgroup$ Commented Feb 27, 2021 at 0:39
  • $\begingroup$ I found the solutions to $\sqrt{i}$ online. $\endgroup$ Commented Feb 27, 2021 at 0:39
  • $\begingroup$ try writing $z$ as $r(\cos{\theta}+i\sin{\theta})$. Because $|z^2|=|z|^2=|i|=1$, so $r=|z|=1$. $\endgroup$ Commented Feb 27, 2021 at 0:42
  • $\begingroup$ The polar form of a complex number is as $re^{i\phi}$ where $r$ is a positive real number. So writing $-i=ie^{i(\pi+k(2\pi))}$ is not helpful, because $i$ is not positive real. $\endgroup$ Commented Feb 27, 2021 at 0:44
  • $\begingroup$ The magnitude of $-i$ is $1$. The phase angle is $3\pi/2$. $\endgroup$ Commented Feb 27, 2021 at 0:45

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Here it is another solution for the sake of curiosity.

Let $z = x + yi$, where $x,y \in \mathbb{R}$. Then we have the following equation:

\begin{align*} z^{2} + i = 0 & \Longleftrightarrow (x+yi)^{2} = x^{2} - y^{2} + 2xyi = -i\\\\ & \Longleftrightarrow \begin{cases} x^{2} - y^{2} = 0\\\\ 2xy = -1 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x = -y\\\\ 2y^{2} = 1\\ \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x = -\dfrac{1}{\sqrt{2}}\\\\ y = +\dfrac{1}{\sqrt{2}} \end{cases}; \begin{cases} x = +\dfrac{1}{\sqrt{2}}\\\\ y = -\dfrac{1}{\sqrt{2}} \end{cases} \end{align*}

Hopefully this helps!

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  • $\begingroup$ I actually thought of this and tried to solve it that way. However, when I tried to solve $x^2=y^2 \rightarrow x=\pm y$ and making the substitutions, $x=y=\pm \sqrt{\frac{1}{2}} i$. In the end, I ended up with 4 different equations for x's and y's, so I was not sure which combination of the x's and y's would give me the answer $\endgroup$ Commented Feb 27, 2021 at 1:12
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    $\begingroup$ This is a good question. The thing is that $x = y$ implies that $2y^{2} = -1$, which is not possible because $y\in\mathbb{R}$. Consequently, it remains just one possibility, that is to say, $x = -y$. If you still have any questions, please let me know. $\endgroup$ Commented Feb 27, 2021 at 1:15
  • $\begingroup$ Ohh right. I completely forgot that $x,y \in \mathbb{R}$ $\endgroup$ Commented Feb 27, 2021 at 1:16
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    $\begingroup$ It's so easy to just jump straight to $re^{i\theta}$ on a question like this, thanks for a very elegant alternative! $\endgroup$ Commented Feb 27, 2021 at 1:22
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    $\begingroup$ Remember that $z = x + yi$. Consequently, the solution set is given by $$S = \left\{-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right\}$$ If you find my answer useful, upvote it. $\endgroup$ Commented Mar 1, 2021 at 3:29
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Here's the solution to $z^2 = i$. Give $z^2 + i = 0$ a try after looking through this.

You have the right idea in using $ z = re^{i\theta} $. First, note that we can choose $r = 1$ because $|re^{i\theta}| = |r|$, and $|i| = 1$. Then, substitute:

$$ z^2 = i$$ $$ (e^{i\theta})^2 = i $$ $$ e^{i(2\theta)} = i $$ $$ \cos(2\theta) + i\sin(2\theta) = i . $$

Comparing the real parts of the left and right-hand sides, we have

$$ \cos(2\theta) = 0 $$ $$ \Rightarrow 2\theta = \frac{\pi}{2} + \pi k $$ $$ \Rightarrow \theta = \frac{\pi}{4} + \frac{\pi}{2}k $$

Now comparing the imaginary parts, we have

$$ \sin(2\theta) = 1 $$ $$ \Rightarrow 2\theta = \frac{\pi}{2} + 2\pi k $$ $$ \Rightarrow \theta = \frac{\pi}{4} + \pi k $$

The two solutions have an intersection of $ \theta = \frac{\pi}{4} + \pi k $. In one rotation from $\theta = 0$ to $\theta = 2\pi$, that gives us two unique solutions:

$$ \theta = \frac{\pi}{4} \ \text{or} \ \theta = \frac{5\pi}{4} .$$

Substituting these back into $z = e^{i\theta} = \cos(\theta) + i\sin(\theta) $, we get the solutions you pointed out:

$$ \theta = \frac{\pi}{4} \Rightarrow z = \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i , $$ $$ \theta = \frac{5\pi}{4} \Rightarrow z = -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i . $$

Now, for $z^2 = -i$, the solution should be a rotation of $\frac{\pi}{2}$ from the solution we got here, i.e., you should get to $\theta = \frac{3\pi}{4} \ \text{or} \ \frac{7\pi}{4}$.

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  • $\begingroup$ Ahhh. Thats very smart. Thanks! $\endgroup$ Commented Feb 27, 2021 at 0:59
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    $\begingroup$ You're very welcome! Check out roots of unity, it's the same intuition with a rotation. $\endgroup$ Commented Feb 27, 2021 at 1:04
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I was hoping to find a similar solution for z=−i−−√ but my attempt gives me z=±i32

I don't see why.

$z = x+yi$ and $z^2 = (x^2 -y^2) + 2xyi = -i$ so $x^2-y^2 = 1$ and $2xy = -1$ so $x^2 =y^2$ and $x = \pm |y|$ but $2xy =-1$ is negative os $x = -y$ and so $2xy=-1\implies -2y^2 = 1\implies y =\pm \frac 1{\sqrt 2}$ and $x = \mp \frac 1{\sqrt 2}$. and $z = \pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2} i$.

Which shouldn't surprise us as if $(\pm \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2} i)^2 = i$ then $i(\pm \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2} i) = \pm \frac 1{\sqrt 2}i \mp \frac 1{\sqrt 2} =\mp \frac 1{\sqrt 2} \pm \frac 1{\sqrt 2}$ when squared should be $i^2*i = -i$.

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Using polar coordidates $-i = 0 + (-1)i = \cos \frac {3\pi}2 +\sin(\frac {3\pi}2)i = e^{(\frac {3\pi}2 + 2k\pi)i}$ and so the square roots of $-i$ are

$e^{(\frac {3\pi}4 + k\pi)i} = \cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases})+\sin(\cos(\begin{cases}\frac {3\pi}4\\\frac {7\pi}4\end{cases}))i =\pm \frac 1{\sqrt 2} \mp \frac 1{\sqrt 2}i$

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