# Complex numbers equation: $z^4 = -16$

Probably dumb question, but I would like to ask it anyway.

I was to solve this equation:

$z^4 = -16$

At first glance, the way to solve it would be (as any other equation):

$z^2 = \sqrt{-1} * 4 \lor z^2 = -\sqrt{-1} * 4$

$z^2 = 4i \lor z^2 = -4i$

$z = 2\sqrt{i} \lor z = -2\sqrt{i} \lor z = 2i\sqrt{i} \lor z = -2i\sqrt{i}$

However, in suggested answers, there are $z_1, z_2, z_3$ and $z_4$ found using trigonometric formulas. Is there something particulary wrong in my solution and the only proper one uses trigonometric form of complex number?

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When you write $\sqrt{i}$, what is that? Angle $\pi/4$ with the positive real axis. Presumably something expressed using trigonometric functions... – GEdgar Feb 13 '12 at 22:13
By solve they mean that one should give the solutions in the form $a+bi$ where $a$ and $b$ are real. – André Nicolas Feb 13 '12 at 22:13
Oh no, there's nothing wrong with what you did. I believe that the suggested solutions just made use of the fourth roots of unity which are $w_k = (\cos(k \pi/2) + i \sin(k \pi/2))$ for $k=1,2,3,4$. These $w_k$ have the property that $w_k^4=1$, and can be used to construct fourth roots. – Eric Haengel Feb 13 '12 at 22:18
So my way of solving this one is generally right, but I would just have to get rid of $\sqrt{i}$ from it? I mean, I (now) know it would be faster to just use those formulas, but the problem with my solution is that it is only half of it, is it what you mean? – Griwes Feb 13 '12 at 22:20
@Griwes: As I mentioned, you solved the problem under your meaning of solve. That is probably not the intended meaning. The two square roots of $i$ are $\pm\frac{1}{\sqrt{2}}(1+i)$. – André Nicolas Feb 13 '12 at 22:40

Say we want the two square roots of $4i$. The angle between the positive real axis and the line on which $4i$ lies is $90^\circ$ or $-270^\circ$. For the square root, you want half that angle, since angles are added when complex numbers are multiplied. It's enough to find one of the two square roots, since after that we can just write "$\pm$" in front of it. Half of $90^\circ$ is $45^\circ$, so we want $a + ai$, i.e. the real and imaginary parts are equal to each other when the angle is $45^\circ$. The absolute value $|a+ai$ is $\sqrt{a^2+a^2}= a\sqrt{2}$ ($a$ is positive because of what we said earlier). We want the absolute value to be $2$ so that the absolute value of the square will be $4$. So we need $a\sqrt{2} = 2$. Therefore $a=\sqrt{2}$.

The two square roots of $4i$ are therefore $\pm(\sqrt{2} + i\sqrt{2})$.

Use a similar method for $-4i$.

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Short way: polar coordinates. $$z^4=-16=(-1)2^4=2^4e^{i\pi}=2^4e^{i\pi+2k\pi i}=2^4e^{i\pi(2k+1)}\;,\;\;k\in\mathbb Z\;\;$$ from which you get $$z=2e^{i\frac\pi 4(2k+1)}\;,\;\;k\in\mathbb Z$$ which assumes only $4$ different values as $k\in\mathbb Z$, so we can write $$z_k=2e^{i\frac\pi 4(2k+1)}\;,\;\;k=0,1,2,3$$ that are the four $4$-th roots of $-16$.

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Therefore, if you continue on from @Joe's proposed solution, you should get

Z1=$2cis(\pi/4)=\sqrt2+\sqrt2i$;

Z2=$2cis(3\pi/4)=-\sqrt2+\sqrt2i$;

Z3=$2cis(5\pi/4)=-\sqrt2-\sqrt2i$;

Z4=$2cis(7\pi/4)=\sqrt2-\sqrt2i$.

If you check, it also makes sense, because complex numbers come in conjugate pairs. Hope it helps.

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We have $$z^4 + 16 = (z^2 + 4)^2 - 8z^2 = (z^2 - 2\sqrt{2} z + 4)(z^2 + 2\sqrt{2} z + 4).$$ Anotherway $$z^2 - 2\sqrt{2} z + 4 = (z-\sqrt{2})^2 + 2 = (z-\sqrt{2})^2 -(\sqrt{2} i)^2 = (z-\sqrt{2} + \sqrt{2} i)(z-\sqrt{2} + \sqrt{2} i)).$$ The same for $z^2 + 2\sqrt{2} z + 4$

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