work out the $\sqrt{2i-1}?$
$2i-1=(a+bi)^2$
$a^2+2abi-b^2$
$a^2-b^2=-1$
$2ab=2$
$a^2=b^{-2}$
$b^{-2}-b^2=-1$
$-b^{4}+1=-1$
$b^4=2$
$b=\sqrt[4]{2}$
Can we solve $\sqrt{2i-1}$ in another way?
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Do you know any field/Galois theory? The tools it provides are really nice for problems like these. Other than that, I don't see why this is a bad approach. – Rushabh Mehta May 15 '19 at 02:14
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No i don't, can you show it – May 15 '19 at 02:15
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It's not something really that I can show just as is. You need a bit of background to take on that material. In that case, what you've done is more than fine. – Rushabh Mehta May 15 '19 at 02:17
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1You made a mistake going from $b^{-2}-b^2=-1$ to $-b^4+1=-1$. It should be $-b^4+1=-b^2$. – Somos May 15 '19 at 02:28
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2Please search up how to convert a complex number to its polar form and the DeMoivre's Theorem. – minori minus May 15 '19 at 02:28
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2The square root has 2 solutions. You need to get them both. – herb steinberg May 15 '19 at 02:30
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@DonThousand could you please write an answer with the tools you mentioned? The answer will be a good starting point to learn more about the tools (for me at least) – user1952500 May 15 '19 at 02:35
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1@user1952500 At some point, sure. – Rushabh Mehta May 15 '19 at 04:31
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https://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number – Hans Lundmark May 15 '19 at 06:04
1 Answers
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I am using polar coordinates.
In short, the polar or trigonometric form of the complex number $z=a+bi$ is $z=r(\cos\theta + i\sin\theta)$. Here, $r=\sqrt{a^2+b^2}$, and $\theta = \tan^{-1}{\frac ba}$ for $a>0$ or $\theta = \tan^{-1}{\frac ba + \pi}$ for $a<0$.
Converting $2i-1$ to polar form, you get $z=\sqrt 5(\cos(\tan^{-1}-2 + \pi) + i\sin(\tan^{-1}-2 + \pi)).$
DeMoivre's theorem states that for a positive integer $p$, the complex number $r(\cos\theta + i\sin\theta)$ has $p$ distinct $p$th roots. They are found by $$r^{\frac1p}\left(\cos\left(\frac{\theta+2\pi n}p\right)+i\sin\left(\frac{\theta+2\pi n}p\right)\right)$$$n = 0,1,2,...,p-1$
Here, $n=2$ (square root). I think you know how to proceed. Remember to calculate for both roots.
minori minus
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