I am unsure on how to do so with I under the square root. I can do so by just simply squaring both sides but I am trying to do so without using this method (need q for part of an equation).
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Try to write it using exponentials – ℋolo Aug 07 '18 at 15:08
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The square root of a complex number is not uniquely defined. When you write $\sqrt{a+ib}$, what do you mean? The usual interpretation is to interpret the square root in terms of the principle branch of the logarithm... – Xander Henderson Aug 07 '18 at 15:09
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1See THIS ANSWER, which provides a full development. And see Method 2 of THIS ANSWER, which circumvents the use of polar coordinates. – Mark Viola Aug 07 '18 at 16:26
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Hint: Write the number under the square root in its polar form. – Pjonin Aug 07 '18 at 15:09
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Hint: Converting to polar coordinates may help:
$$q = \sqrt{a+ib} = \sqrt{re^{i\theta}} = \sqrt{r}e^{i\theta/2}$$
Use the conversions $r=\sqrt{a^2+b^2}$ and $\theta = \tan^{-1}(b/a)$. (For the angle, you'll need to take the quadrant of the point into account.)
John
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It's not correct that $\theta = \tan^{-1}(b/a)$. This only works if $a>0$. Otherwise you will have to add/subtract $\pi$ (see e.g. https://en.wikipedia.org/wiki/Atan2#Definition_and_computation ) – Winther Aug 07 '18 at 15:58
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@Winther Hi Hans. Perhaps John is interpreting $\arctan(b/a)$ as $\arctan2(b,a)$. – Mark Viola Aug 07 '18 at 16:22
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See THIS ANSWER, which provides a more precise development. And see Method 2 of THIS ONE, which circumvents using polar coordinates. – Mark Viola Aug 07 '18 at 16:26
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@Winther Thanks; I edited to make it clear that you have to pay attention to the quadrant. – John Aug 07 '18 at 17:57