Find all complex numbers that satisfies this equation
$(z - 6 + i)^3 = -27.$
I found one of them being $ z = 3 - i $
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2We usually write $i$ lower case. – Gregory Grant May 05 '15 at 14:40
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2There are three cube roots of $-27$ so there have to be three solutions to this equation. Do you know how to find the three cube roots? – Gregory Grant May 05 '15 at 14:41
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@GregoryGrant i just changed it. – chosenman May 05 '15 at 14:41
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@GregoryGrant not really – chosenman May 05 '15 at 14:43
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2Hint: If $\zeta=\frac{1}{2}+\frac{\sqrt{3}}{2}i$, then $\zeta^3=-1$ and $(\zeta^2)^3=-1$. – Gregory Grant May 05 '15 at 14:43
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$-27 =\left(3e^{-i\pi/3}\right)^3 =\left(3e^{i\pi/3}\right)^3 =\left(3e^{i\pi}\right)^3$ – robjohn May 05 '15 at 14:43
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2Note that $\zeta^2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ – Gregory Grant May 05 '15 at 14:44
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2If you know how to do that (using long division), you can now factor the polynomial and get a quadratic equation that you should be able to solve. – May 05 '15 at 14:45
3 Answers
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Let $z-6+i=\omega$, then your equation change to find the solution of equation $\omega^3=-27$. this equation have the solution's $z_k=3(cos(\frac{\pi+2k\pi}{3})+isin(\frac{\pi+2k\pi}{3}))$ which $k=0,1,2$. Finally: $z=6-i+z_k$, are all solution's.
hamid kamali
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$-27=27e^{{\pi}i+2k\pi}$
$\sqrt[3]{27e^{{\pi}i+2k\pi}}=3e^{\frac{1}{3}\pi i+\frac{2}{3}k\pi}=z-6+i$
$z=3e^{\frac{1}{3}\pi i+\frac{2}{3}k\pi}+6-i, \forall k\in \mathbb{Z}$
Dasherman
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