2
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$ (z^2-1)^2 = 4 \iff $$z_1 = 3 $ and $ z_2=-1$

$arg(z_1)= 0 , arg(z_2) = \pi$

$$ z_1 = \sqrt{3} \left(\cos\left(\frac{2\pi k}{2}\right) + i\sin\left(\frac{2\pi k}{2}\right)\right)$$

$$z_2 = i \left(\cos\left(\frac{2\pi k}{2} +\pi\right) + i\sin\left(\frac{2\pi k}{2} +\pi\right)\right) \mid k= \{ 0,1 \}$$

$$ z = \{ \sqrt{3}, -\sqrt{3}, i, -i \}$$

Is this correct?

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  • 1
    $\begingroup$ You mean $z_1^2=3$ and so on $\endgroup$ – Andrea Mori Jan 19 '15 at 15:21
  • $\begingroup$ Yes, it seems right. $\endgroup$ – Hardey Pandya Jan 19 '15 at 15:26
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$(z^2-1)^2=4\iff$

$z^2-1=2,-2\iff$

$z^2=3,-1\iff$

$z=\sqrt{3},-\sqrt{3},\sqrt{-1},-\sqrt{-1}\iff$

$z=\sqrt{3},-\sqrt{3},i,-i$

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  • $\begingroup$ This is nice and simple. I hardly can believe this is "legal". Can you really handle complex powers like with real-numbers?? $\endgroup$ – user1511417 Jan 19 '15 at 15:46
  • $\begingroup$ @user1511417: Yes. $\endgroup$ – barak manos Jan 19 '15 at 15:47
  • $\begingroup$ Caveat: When you take the $n$th root of both sides, you will have $n$ solutions, some of which you will have ignored with real numbers. $\endgroup$ – GFauxPas Jan 19 '15 at 15:49
  • $\begingroup$ @GFauxPas: True, but in the case above I have not ignored any such solution. $\endgroup$ – barak manos Jan 19 '15 at 15:50
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    $\begingroup$ @user1511417 Google "roots of unity". Have fun! I recommend the (e-)book by Brown and Churchill $\endgroup$ – GFauxPas Jan 19 '15 at 16:10
2
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$(z^2-1)^2=4\Leftrightarrow (z^2-1)^2-2^2=0\Leftrightarrow (z^2-1-2)(z^2-1+2)=0\Leftrightarrow(z^2-3)(z^2+1)=0\Leftrightarrow (z-\sqrt 3)(z+\sqrt 3)(z-i)(z+i)=0$ It follows that $z\in\{\sqrt 3, -\sqrt 3, i,-i\}$ .

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