Question:
Solve $z^5 =32$ (including complex solutions)
My Working :
$$ z^5=32 $$ $$ z^5=32cis(0+2k\pi) $$ $$ z=2cis(0+\frac{2k\pi}5) $$ I then went through all the solutions for $k=0\to4$ which gave me the solutions of: $$ 2,\space2cis(0+\frac{2\pi}5),\space2cis(0+\frac{4\pi}5),\space2cis(0+\frac{6\pi}5),\space2cis(0+\frac{8\pi}5) $$
Could someone confirm for me that this is correct, and if not where I have gone wrong?
Thanks!
$\endgroup$ 62 Answers
$\begingroup$The complex solutions to your problem is $2e^{\dfrac{i2k\pi}{5}}$ where $k=0,1,2,3,4$
$\endgroup$ 3 $\begingroup$As others have noted, the solutions are indeed $$z_k = 2{\rm e}^{2k\pi{\rm i}/5}=2{\rm cis}\Bigl({2k\pi\over 5}\Bigr)=2\Bigl(\cos\Bigl({2k\pi\over 5}\Bigr)+{\rm i}\sin\Bigl({2k\pi\over 5}\Bigr)\Bigr),\qquad k=0,1,2,3,4$$ But as the angles ${2\pi k \over 5}=k\cdot 72^\circ$ have explicit values for cosine and sine, the solutions can also be written as $$z_1 = 2\Bigl({\sqrt 5 - 1\over 4}+{\rm i}{\sqrt{10+2\sqrt 5} \over 4}\Bigr)$$ $$z_2 = 2\Bigl(-{\sqrt 5 + 1 \over 4}+{\rm i}{\sqrt{10-2\sqrt 5}\over 4}\Bigr) $$ together with $z_0 = 2$, $z_3 = \bar z_2$, and $z_4 = \bar z_1$.
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