Find all possible values $\sqrt{i}+\sqrt{-i}$

Question

Find all possible values of $\sqrt{i} + \sqrt{-i}$. Here $i= \sqrt{-1}$.

My solution which didn't work.

\begin{eqnarray}a+ib &=& \sqrt{i}+\sqrt{-i}\\ &=& \sqrt{i}+\sqrt{i^2i}\\ &=& \sqrt{i}+\sqrt{i}i\end{eqnarray} \begin{eqnarray}(a+ib)^2 &=& \left(\sqrt{i}+\sqrt{i}i\right)^2\\a^2-b^2+2abi &=& -1-i-2i\\ &=& -1-3i\end{eqnarray} \begin{cases}a^2-b^2=-1\\2ab=-3\end{cases} \begin{cases}a^2-b^2=-1\\4a^2b^2=9\end{cases} \begin{eqnarray}\left(a^2+b^2\right)^2 &=& \left(a^2-b^2\right)^2+4a^2b^2\\ &=& 1+9\\ &=& 10\end{eqnarray} \begin{cases}a^2-b^2=-1\\a^2+b^2=\sqrt{10}\end{cases} \begin{eqnarray}2a^2 &=& \sqrt{10}-1\\a^2=\frac{\sqrt{10}-1}2\\ a &=& \sqrt{\frac{\sqrt{10}-1}2}\\b &=& \sqrt{\frac{\sqrt{10}+1}2}\end{eqnarray}

Answer 1

I actually liked your approach! Unfortunately, a misconception and a miscalculation took you off the rails.

The misconception comes from the fact that we're used to thinking of $\sqrt{\alpha}$ as a single number, which is sensible as long as we're only dealing with non-negative real numbers. However, beyond that, we need to consider it to indicate one of up to two different numbers, which we can't narrow down further without more information. Your book doesn't help this by saying things like "$i=\sqrt{-1},$" unfortunately.

In particular, we must keep this in mind for any complex number $z$: $$\sqrt{z^2}=\pm z.\tag{$\star$}$$

\begin{eqnarray}a+ib &=& \sqrt{i}+\sqrt{-i}\\ &=& \sqrt{i}+\sqrt{i^2i}\\ &=& \sqrt{i}+\sqrt{i}i\end{eqnarray}

We've got to be a little careful trying to use the property $$\sqrt{\alpha\beta}=\sqrt{\alpha}\sqrt{\beta}$$ when $\alpha$ and $\beta$ aren't nonnegative reals, because of $(\star).$ We should instead have $$a+ib=\sqrt{i}\pm\sqrt{i}i.$$

\begin{eqnarray}(a+ib)^2 &=& \left(\sqrt{i}+\sqrt{i}i\right)^2\\a^2-b^2+abi &=& -1-i-2i\\ &=& -1-3i\end{eqnarray}

I'm not sure what happened on the right-hand side. It should end up being $-2,$ which of course changes everything! From there, you'll see that $a=0$ and $b=\pm\sqrt2,$ giving us $\pm\sqrt{2}i$ as two of our possible solutions. Considering also the case $a+ib=\sqrt{i}-\sqrt{i}i,$ we obtain instead $b=0$ and $a=\pm\sqrt 2,$ which gives us the other two solutions: $\pm\sqrt{2}$!

However, even if this had been correct, some more things went wrong on the way, as well.

\begin{cases}a^2-b^2=-1\\2ab=-3\end{cases} \begin{cases}a^2-b^2=-1\\4a^2b^2=9\end{cases} \begin{eqnarray}\left(a^2+b^2\right)^2 &=& \left(a^2-b^2\right)^2+4a^2b^2\\ &=& 1+9\\ &=& 10\end{eqnarray} \begin{cases}a^2-b^2=-1\\a^2+b^2=\sqrt{10}\end{cases}

Fine so far, but only because we can't have $a^2+b^2=-\sqrt{10}$ when $a,b$ are real!

\begin{eqnarray}2a^2 &=& \sqrt{10}-1\\a^2 &=&\frac{\sqrt{10}-1}2\\ a &=& \sqrt{\frac{\sqrt{10}-1}2}\end{eqnarray}

Here, again, we should have used $(\star)$ to instead have obtained $$a=\pm\sqrt{\frac{\sqrt{10}-1}2},$$ and similarly, we'd have had $$b=\pm\sqrt{\frac{\sqrt{10}+1}2}.$$

Answer 2

Hint

Express $$i=e^{i{\pi\over2}}\\-i=e^{i{3\pi\over2}}$$then $$x_1=r_1e^{i\theta_1} , x_1^2=i\\x_2=r_2e^{i\theta_2} , x^2_2=-i$$what are the possible values of $x_1+x_2$?

Answer 3

Hint: $i=\frac{1}{2}(1+i)^2$ and $-i=\frac{1}{2}(1-i)^2$.

Answer 4

Hint:

If $(a+bi)^2=\alpha+\beta i$, then: $$a=\pm\sqrt{\frac{\alpha\pm\sqrt{\alpha^2+\beta^2}}{2}}, b=\pm\sqrt{\frac{-\alpha\pm\sqrt{\alpha^2+\beta^2}}{2}}$$

Can you use this to find $\sqrt i$ and $\sqrt{-i}$ in the form $a+bi$?

If you're wondering how I got these expressions, it's a nice bit of algebra from the first statement. Try it yourself!

Answer 5

@Cameron Buie has written an excellent answer, but adding my two cents to it, I would like to elaborate this:

\begin{eqnarray}(a+ib)^2 &=& \left(\sqrt{i}+\sqrt{i}i\right)^2\\a^2-b^2+abi &=& -1-i-2i\\ &=& -1-3i\end{eqnarray}

The part where it wrong was in squaring the $\sqrt{i}$ it must be i and not -1 also, $\sqrt{i}.i$ squared would be -i. And evaluating the third term, it would be a nice -2 that satisfies the answer that Cameron Buie points out in their answer.

P.S. This was supposed to be a comment/suggestion but since I don't have the required reputation, I wrote it as a separate answer.