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I'm trying to solve it by using its polar form, but then I get

$$ \begin{align*} z^4 &= (\rho e^{i\phi})^4 = \rho^4 e^{4i\phi}\\ &= -4 = -4 e^{0i}\\ \end{align*} $$ From the definition of equality of complex numbers, $\rho^4 = -4$ and $4\phi = 0 + 2\pi k$ for some $k \in \mathbb{Z}$.

This would mean $\rho = \sqrt{-2}$ and $\phi = \pi k / 2$. I have no idea how to interpret that imaginary radius, and Wolfram Alpha says the angle is $\pi k / 4$. Should this equation be solved using this method? Am I missing some step, or are my calculations incorrect?

I've already read Solve $z^4+1=0$ algebraically, but I want to know whether this equation is solvable using this method or another method should be used.

Community
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user1002327
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    The radius should not be negative - account for this negativity with the angle instead (remember, $e^{i\pi} = -1$) – Stahl Nov 28 '13 at 01:33

4 Answers4

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The modulus of a nonzero complex number is always positive. So from your equation you get $$ \rho^4\,e^{4i\phi}=4(-1)=4e^{-i\pi}. $$ From there you get $\rho^4=4$ and $4\phi=2k\pi-\pi$ for $k$ integer. So $$ \rho=4^{1/4}=\sqrt2,\ \ \phi=\frac{k\pi}2-\frac\pi4=\frac{(2k-1)\pi}4,\ k=1,2,3,4. $$ (adding multiplies of $4$ to $k$ yields multiples of $2\pi$ that will not affect $\phi$)

Martin Argerami
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  • That $e^{-i \pi} = 1$ is exactly what I was missing. I went and tried to blindly solve it without taking that into account. Thank you and Stahl for pointing it out. – user1002327 Nov 28 '13 at 01:43
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You want to find the fourth roots of $-4$, namely $z=(-4)^{\frac{1}{4}}$.

Here, $\theta=\pi$ and $r=|z|=4$.

Hence, by De'Moivre we have $z=\sqrt{2}(\cos(\frac{\pi+2\pi k}{4})+i\sin(\frac{\pi+2\pi k}{4}))$ for $k=0,1,2,3$.

In polar we have, $z=e^{\frac{(\pi+2\pi k)}{4}i}$ for $k=0,1,2,3$.

Mr.Fry
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Something that we all never learned in high school, but should have, is the amazing factorization $X^4+4=(X^2+2X+2)(X^2-2X+2)$. With this, the total factorization is easy.

Lubin
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Using that $x^2-y^2 = (x+y)(x-y)$ and $x^2+y^2 = (x+iy)(x-iy)$, we get

\begin{align} z^4+4 &= (z^2+2i)(z^2-2i)\\ &= (z+i\sqrt{2i})(z-i\sqrt{2i})(z+\sqrt{2i})(z-\sqrt{2i})\\ &= (z-1+i)(z+1-i)(z+1+i)(z-1-i) \end{align}

where the last equality is due to $(1+i)^2 = 2i$.

Ennar
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