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Here's another one of the problems I'm having trouble with. Below are two statements and the book is asking me to tell for each whether it is true or false:

  1. All the solutions to the equation $\ \ z^3=i\ \ $ are:

$z_1=-i\ ,\ z_2=\frac{\sqrt3}{2}+\frac{i}{2}\ ,\ z_3=-\frac{\sqrt3}{2}+\frac{i}{2}$

  1. All the solutions to the equation $\ \ z^2=i-\sqrt3\ \ $ are:

$z_1=\sqrt2\left( \cos\frac{\pi}{3}+i\sin\frac{\pi}{3} \right)\ ,\ z_2=\sqrt2\left( \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} \right)$

I know that the three answers listed for the first one are indeed roots of $i$ but I'm not so sure about the second statement.

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    $\begingroup$ $z^2+(\sqrt{3}-i)$ has exactly $2$ complex roots by the Fundamental Theorem of Algebra. Clearly $z_1\neq z_2$, so all you would have to do is to actually compute $z_1^2$ and $z_2^2$ and verify if they, indeed, satisfy the equation. $\endgroup$
    – Guest
    Dec 18, 2015 at 20:16
  • $\begingroup$ For the first: $i^3 = -i \neq i$. $\endgroup$
    – Théophile
    Dec 18, 2015 at 20:29
  • $\begingroup$ Oh I think I copied the first part wrong - $z_1$ should be $-i$. Sorry about that. $\endgroup$
    – Or Bairey-Sehayek
    Dec 18, 2015 at 20:32

1 Answer 1

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$$z^2=i-\sqrt{3}\Longleftrightarrow$$ $$z^2=|i-\sqrt{3}|e^{\arg(i-\sqrt{3})i}\Longleftrightarrow$$ $$z^2=2e^{\frac{5\pi i}{6}}\Longleftrightarrow$$ $$z=\left(2e^{\left(2\pi k+\frac{5\pi}{6}\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$ $$z=\sqrt{2}e^{\frac{1}{2}\left(2\pi k+\frac{5\pi}{6}\right)i}$$

With $k\in\mathbb{Z}$ and $k:0-1$

So the solutions are:

$$z_0=\sqrt{2}e^{\frac{1}{2}\left(2\pi\cdot0+\frac{5\pi}{6}\right)i}=\sqrt{2}e^{\frac{5\pi i}{12}}=\sqrt{2}\left(\cos\left(\frac{5\pi}{12}\right)+\sin\left(\frac{5\pi}{12}\right)i\right)$$ $$z_1=\sqrt{2}e^{\frac{1}{2}\left(2\pi\cdot1+\frac{5\pi}{6}\right)i}=\sqrt{2}e^{-\frac{7\pi i}{12}}=\sqrt{2}\left(\cos\left(-\frac{7\pi}{12}\right)+\sin\left(-\frac{7\pi}{12}\right)i\right)$$

Notice with $a,b\in\mathbb{R}$: $$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|\left(\cos(\arg(a+bi))+\sin(\arg(a+bi))i\right)$$

With $e$ is the base of the natural logarithm.

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  • $\begingroup$ I think what you used here is polar notation and we never learned that in class. What does the $e$ mean here? If I'm understanding correctly, $e^{\frac{5\pi i}{12}}$ means $\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}$ - is that correct? $\endgroup$
    – Or Bairey-Sehayek
    Dec 18, 2015 at 20:20
  • $\begingroup$ @OrBairey-Sehayek Look to my edit, I hope it is understandable now $\endgroup$
    – Jan Eerland
    Dec 18, 2015 at 20:24
  • $\begingroup$ Ah okay, thanks. $\endgroup$
    – Or Bairey-Sehayek
    Dec 18, 2015 at 20:25
  • $\begingroup$ @OrBairey-Sehayek You're more than welcome! $\endgroup$
    – Jan Eerland
    Dec 18, 2015 at 20:26

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