$z^4+i =0$ How to factorize it and find the roots

Question

$z^4+i=0$ where $z \in \mathbb{C}$ should have $4$ roots by F $\theta$ C.

However I plugged this into Mathematica and it couldn't factorize it or find the roots. I tried myself, but by looking at the powers of $i$ I see:

$$i^2 = -1$$ $$i^3 = -i$$ $$i^4 = 1$$ $$i^5=i$$

that the fourth square root of $-i$ doesn't exists, as we are back where we started.

How do I factor this? And what are the roots?

Answer 1

We have $$ z^4 + i = 0 \iff \\ z^4 = -i \quad (*) $$ We now move to polar coordinates, expressing $z$ as $$ z = r e^{i\phi} $$ with radius $r \in [0, \infty)$ and angle $\phi \in [0, 2\pi)$. Then $(*)$ turns into $$ (r e^{i\phi})^4 = r^4 e^{4i \phi} = e^{i (3 \pi/2 + 2\pi k)} \quad (k \in \mathbb{Z}) $$ where we added the cyclic repetition of $e^{i\phi}$ to the right hand side.

Comparing both sides we have $$ r = 1 \\ 4\phi = \frac{3\pi}{2} + 2\pi k $$ which means $$ \phi = \frac{3\pi}{8} + \frac{k \pi}{2} = \frac{(3+4k)\pi}{8} $$ where $$ \frac{3\pi}{8}, \frac{7\pi}{8},\frac{11\pi}{8},\frac{15\pi}{8} $$ are in $[0, 2\pi) = [0, 16\pi/8)$.

This gives the roots $$ z \in \left\{ e^{3\pi i/8}, e^{7\pi i/8}, e^{11\pi i/8}, e^{15\pi i/8} \right\} $$

(Large version)

Answer 2

Hint: You need to find $$ z=\sqrt[4]{-i}=\sqrt[4]{e^{i(3\pi/2+2\pi k)}},\quad k\in\mathbf Z, $$ use the properties of the exponential function and see for which $k$ you'll get different roots.

Answer 3

$$ \large \begin{align} z^4+i&=0\\ z^4&=-i\\ &=e^{i(2n-\frac 12)\pi}\\ z&=e^{i(\frac n2-\frac 18)\pi}\\ &=e^{-i\frac {\pi} 8},\; e^{i\frac {3\pi} 8},\; e^{i\frac {7\pi} 8}\; e^{i\frac {11\pi} 8} \end{align}$$

Answer 4

Write $-i=e^{3\pi i/2},e^{7\pi i/2},e^{11\pi i/2},e^{15\pi i/2}$ (i.e go around the circle in the complex plane a few times. Then take the $4^{th}$ root. Note that all angles should be less than $2\pi$

Answer 5

Since $z^4=-i$, you're looking for the fourth roots of $-i$, which look like these points on the unit circle on an Argand diagram:

Note that complex $n$th roots of a number are always separated like this by $2\pi/n$ radians, so that when raised to the appropriate power they are "separated" by $2\pi \equiv$ coincident. Since $-i$ has an argument of $3\pi/2$, the "base" fourth root has an argument of $3\pi/8$.

Answer 6

Let us write the equation as $$z^4-a^4=0$$ for some $a$ to be determined.

We can factor as

$$(z^4-a^4)=(z^2-a^2)(z^2+a^2)$$ and with a little more effort,

$$(z-a)(z+a)(z-ia)(z+ia).$$

So the roots are of the form $a,ia,-a,-ia$, and we also have that $a^4=-i$. Without knowing how to extract the roots of a complex number, you can already write

$$(z-\sqrt[4]{-i})(z-i\sqrt[4]{-i})(z+\sqrt[4]{-i})(z+i\sqrt[4]{-i}).$$

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To go further, you can use the polar form. Remember that the modulus of a product is the product of the moduli, and the argument of a product is the sum of the arguments.

Then the modulus of a fourth power is the fourth power of the modulus ($-i\to1$ in our case) and the argument of a fourth power is four times the argument ($-i\to3\pi/2$).

Hence, $$a=e^{i3\pi/8}.$$

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You can also work this out directly, taking the square root twice.

$$(x+iy)^2=x^2-y^2+i2xy=-i$$

gives the system

$$x^2-y^2=0,\\2xy=-1.$$

Choosing $x=y$, we have the solution

$$\frac1{\sqrt2}(1-i).$$

Now let us solve

$$x^2-y^2=\frac1{\sqrt2},\\2xy=-\frac1{\sqrt2},$$ or after multiplication by $x^2$,

$$x^4-x^2y^2=\frac{x^2}{\sqrt2}$$ or

$$x^4-\frac{x^2}{\sqrt2}-\frac18=0.$$

One of the real solutions is

$$x=\frac12\sqrt{2+\sqrt2},$$ with $$y=-\frac1{\sqrt2\sqrt{2+\sqrt2}}=-\frac12\sqrt{2-\sqrt2}.$$

Answer 7

Multiply out $z^4$ when $z = \sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}i$
$$ \left( \sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}i \right)\left( \sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}i \right) =\\ (\sqrt{2}-1) - (\sqrt{2}+1) +2 \sqrt{2-1}i = -2+2i \\ (-2+2i)(-2+2i) = 4 - 4 - 8i = -8i $$ So this $z$ gives something times $-i$, but it is too large by a factor of $\sqrt[4]8$. Thus one solution to $z^4=-i$ is $$ z =\frac{ \sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}i}{\sqrt[4]8} $$ The other three solutions are, of course, this $z$ times $-1, i$ and $-i$, respectively.

How does one guess that original form? The best way I can think of is to use deMoivre's theorem, which leads to having to find the sine and cosine of $67.5^\circ$, which are the cosine and sine of $22.5^\circ$. You can then use the half-angle cosine formula on an angle of $45^\circ$, and that clues you into the presence of a square root of $\sqrt{2}/2 + 1$.