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To solve this question, firstly, I let $-i = e^\frac{3i\pi}{2}$. So when I square root it, I will get $\sqrt {-i} = e^\frac{3i\pi}{4}$. When I convert it to Cartesian equation, I get $\sqrt {-i} = \cos(\frac{3i\pi}{4}) + i\sin(\frac{3i\pi}{4})$ which is $-\frac{1}{\sqrt 2} + \frac{i}{\sqrt2}$. But when I checked with google/wolfram'answer, their answer is $\frac{1}{\sqrt 2} - \frac{i}{\sqrt2}$. So I wonder which part am I wrong??

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    $\begingroup$ You can find a quick mathjax tutorial here. Please use it to format your question. Right now, it is unreadable. $\endgroup$
    – P. Siehr
    Commented Oct 6, 2017 at 7:19
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    $\begingroup$ In general it is a good idea to avoid using the $\sqrt{}$ symbol for the square root function for anything else than non-negative real numbers as there is no preferred answer to the question e.g. what $x$ satisfies $x^2=-i$? $\endgroup$
    – M. Winter
    Commented Oct 6, 2017 at 7:41

3 Answers 3

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Notice that $-\frac{1}{\sqrt2}+\frac{i}{\sqrt2}=-\left(\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right)$

Since you are finding a square root, this can be both positive and negative, your answer is not wrong, it is just another value.

Also, one part you missed out is that with trigonometric functions, you need to add $2k\pi$ where $k$ is an integer. Doing so would enable you to get all the solutions to the equation instead of only $1$.

So $-i=e^{i\left(\frac{3\pi}{2}+2k\pi\right)}$

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  • $\begingroup$ Thanks for helping.. $\endgroup$
    – kevin
    Commented Oct 6, 2017 at 7:26
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When you are asked to find the square root of $-i$ (or the square root of any number), then you are meant to find $x$ such that:

$$x^2 = -i.$$

Now, you know that $-i = e^\frac{3i\pi}{2}$, and hence you must solve this equation:

$$x^2 = e^\frac{3i\pi}{2}.$$

The solutions are $2$:

$$x = \pm \left(e^\frac{3i\pi}{2} \right)^{\frac{1}{2}} = \pm\left(-\frac{1}{\sqrt 2} + \frac{i}{\sqrt2}\right).$$

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  • $\begingroup$ Thanks for helping $\endgroup$
    – kevin
    Commented Oct 6, 2017 at 8:08
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There are two square roots. You only found one of them. Negate it to get the other.

Wolfram Alpha chose to show the other one.

The correct solution is to show both.

Alternatively, represent $-i$ as $$e^{i \left( {\Large{-\frac{\pi}{2}}} \right) } $$ and then divide the angle by $2$, to get $$e^{i \left( {\Large{-\frac{\pi}{4}}} \right) } $$ That gives you the other one, but then, if you choose to do that one first, also negate it, so as to get both square roots.

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  • $\begingroup$ thanks for taking your time to help.. $\endgroup$
    – kevin
    Commented Oct 6, 2017 at 7:26

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