What is the value of $x$ in $x^4+1=0$?

Question

So I was looking through the homepage of Youtube when I came across this video by Blackpenredpen which asked the question$$\text{What is the value of }x\text{ in }x^4+1=0\text{?}$$which I thought that I might be able to solve. Here is my attempt at solving the aforementioned equation:$$x^4=-1$$$$\therefore x^2=\pm i$$$$\therefore x=\pm\dfrac i{\sqrt2}\pm\dfrac1{\sqrt2}$$However, since I have sadly lost the page explaining the expansion of the square root of $i$, here is $\pm\dfrac i{\sqrt2}\pm\dfrac1{\sqrt2}$ squared:$$\left(\dfrac{i\sqrt2+\sqrt2}2\right)^2$$$$=\dfrac{i\sqrt2(i\sqrt2+\sqrt2)+\sqrt2(i\sqrt2+\sqrt2)}4$$$$=\dfrac{4i}4=i$$

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$$\left(\dfrac{i\sqrt2-\sqrt2}2\right)^2$$$$=\dfrac{i\sqrt2(i\sqrt2-\sqrt2)-\sqrt2(i\sqrt2-\sqrt2)}4$$$$=-\dfrac{4i}4=-i$$

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$$\left(\dfrac{-i\sqrt2-\sqrt2}2\right)^2$$$$=\dfrac{-i\sqrt2(-i\sqrt2-\sqrt2)-\sqrt2(-i\sqrt2-\sqrt2)}4$$$$=\dfrac{4i}4=i$$

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$$\left(\dfrac{-i\sqrt2+\sqrt2}2\right)^2$$$$=\dfrac{-i\sqrt2(-i\sqrt2+\sqrt2)+\sqrt2(-i\sqrt2+\sqrt2)}4$$$$=\dfrac{-4i}4=-i$$Therefore the solutions to $x^4+1=0$ are:$$\pm\dfrac i{\sqrt2}\pm\dfrac1{\sqrt2}$$

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My question

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Is the solution that I have achieved correct, or what could I do to attain it/attain it more easily?

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Mistakes I might have made

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1. The only one that I can think of is the square rooting part honestly, but there might be other mistakes.

Answer 1

Complex Epnential

Another easy way to solve would be via Euler's formula or Euler's identities. According to these, the following applies: $-1 = e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i} \wedge k \in \mathbb{Z}$ (this is also known as the polar form of $-1$)

With this we get: $$ \begin{align*} x^{4} + 1 &= 0\\ x^{4} &= -1\\ x^{4} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\ x &= \left( e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i} \right)\\ x &= e^{\frac{\pi + 2 \cdot k \cdot \pi}{4} \cdot i}\\ x &= \cos\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) + \sin\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) \cdot i\\ \end{align*} $$

So we get the general solution: $$\fbox{$ \begin{align*} x &= \cos\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) + \sin\left( \frac{\pi + 2 \cdot k \cdot \pi}{4} \right) \cdot i\\ \end{align*} $}$$

We can do this even mor general: $$ \begin{align*} z^{a} &= -1\\ \left( z \right)^{a} &= -1 \tag{Polar Form}\\ \left( \left| z \right| \cdot e^{\arg\left( z \right) \cdot i} \right)^{a} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\ \left( 1 \cdot e^{\arg\left( z \right) \cdot i} \right)^{a} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\ \left( e^{\arg\left( z \right) \cdot i} \right)^{a} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i}\\ e^{a \cdot \arg\left( z \right) \cdot i} &= e^{\pi \cdot i + 2 \cdot k \cdot \pi \cdot i} \quad\mid\quad \ln\left( \cdot \right)\\ a \cdot \arg\left( z \right) \cdot i &= \pi \cdot i + 2 \cdot k \cdot \pi \cdot i \quad\mid\quad \div \left( a \cdot i \right)\\ \arg\left( z \right) &= \frac{1}{a} \cdot \left( \pi + 2 \cdot k \cdot \pi \right)\\ \end{align*} $$ $$\fbox{$ \begin{align*} z &= \cos\left( \frac{1}{a} \cdot \left( \pi + 2 \cdot k \cdot \pi \right) \right) + \sin\left( \frac{1}{a} \cdot \left( \pi + 2 \cdot k \cdot \pi \right) \right) \cdot i\\ \end{align*} $}$$

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Geometry And The Beauty Of The Complex Arguments

With his Euler formula, Leonhard Euler discovered one of the most important relations of complex numbers (that of trigonometry and geometry). So we could also ask ourselves geometrically what it means to take the $n$th root of a number: When rooting with the $n$th root, we divide the argument aka the angle of the number by $n$. Any integer multiple is then the solution.

Here that means: $\arg\left( -1 \right) = \pi \Leftrightarrow \arg\left( \sqrt[4]{-1} \right) = \frac{\pi}{4} + 2 \cdot k \cdot \pi \wedge k \in \mathbb{Z}$

$\sqrt[4]{-1}$" />

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Roots Of Unity

For example, if you square, you get an equation whose solution is the $2 \cdot \left| a \right|$-Roots Of Unity (iff $a \in \mathbb{Z} \setminus \left\{ 0 \right\}$: $$ \begin{align*} z^{a} &= -1 \quad\mid\quad \left( \cdot \right)^{2}\\ \left( z^{a} \right)^{2} &= \left( -1 \right)^{2}\\ z^{2 \cdot a} &= 1\\ \end{align*} $$

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Trigonometric Functions

We know from Euler's Formula ($e^{x \cdot i} = \cos\left( x \right) + \sin\left( x \right) \cdot i$) that we can represent a complex number $z$ in [Polar Form][3] ($z = \left| z \right| \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)$). So, with the help of [De Moivre's formula][4] $\left( \cos\left( x \right) + \sin\left( x \right) \cdot i \right)^{a} = \cos\left( a \cdot x \right) + \sin\left( a \cdot x \right) \cdot i$ we get: $$ \begin{align*} z^{a} &= -1 \tag{polar form}\\ \left( \left| z \right| \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right) \right)^{a} &= -1\\ \left| z \right|^{a} \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)^{a} &= -1\\ \end{align*} $$

We know that $\left| 1^{x} \right| = 1$, so $\left| z \right|^{a} = \left| - 1 \right| = 1$: $$ \begin{align*} \left| z \right|^{a} \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)^{a} &= -1\\ 1 \cdot \left( \cos\left( \arg\left( z \right) \right) + \sin\left( \arg\left( z \right) \right) \cdot i \right)^{a} &= -1 \tag{De Moivre's Formula}\\ \cos\left( a \cdot \arg\left( z \right) \right) + \sin\left( a \cdot \arg\left( z \right) \right) \cdot i &= -1\\ \end{align*} $$

Since the imaginary part of one side of the equation must equal the imaginary part of the other side of the equation, and $-1$ has the imaginary part $0$, $\sin\left( \cdot \right) = 0$ must hold. Thus we get: $$ \begin{align*} \cos\left( a \cdot \arg\left( z \right) \right) + \sin\left( a \cdot \arg\left( z \right) \right) \cdot i &= -1\\ \cos\left( a \cdot \arg\left( z \right) \right) + \sin\left( a \cdot \arg\left( z \right) \right) \cdot i &= -1 + 0 \cdot i\\ \cos\left( a \cdot \arg\left( z \right) \right) &= -1 \quad\mit\quad \arccos\left( \cdot \right)\\ a \cdot \arg\left( z \right) &= \arccos\left( -1 \right) \quad\mit\quad \div a\\ \arg\left( z \right) &= \frac{1}{a} \cdot \arccos\left( -1 \right)\\ \end{align*} $$

$$\fbox{$ \begin{align*} z &= \cos\left( \frac{1}{a} \cdot \arccos\left( -1 \right) \right) + \sin\left( \frac{1}{a} \cdot \arccos\left( -1 \right) \right) \cdot i\\ \end{align*} $}$$

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For another example for equations like this see I'm looking for answer to $z^{e} = -1$.

Answer 2

You can use polar form for the complex number to solve this faster.

$\text{Known}: e^{i\theta} = \cos\theta + i\sin\theta$

You can check for yourself that:

$e^{i\pi} = -1$

Using this :

$\sqrt[\leftroot{-2}\uproot{2}4]{-1} = \pm(e^{i\pi})^{\frac{1}{4}}$

Substituting and solving this will leave with your answer.

Answer 3

Alt. hint: $\;$ note that $x=0$ is not a root, then:

$$ 0 = x^4 + 1 = x^2 \left(x^2 + \frac{1}{x^2}\right) = x^2\left(\left(x+\frac{1}{x}\right)^2 - 2\right) = \left(x^2 - \sqrt{2}\,x + 1\right)\left(x^2 + \sqrt{2}\,x + 1\right) $$

Answer 4

To fill in the single point where you've handwaved a little (or at least worked backwards from the solution), here's a way to find $i^{\frac{1}{2}}$ without trigonometry:

Suppose we have $w = a + bi \in \mathbb{C}$ with $a, b \in \mathbb{R}$ such that $w^2 = i$. Then:

$$\begin{eqnarray}(a + bi)^2 & = & i \\ a^2 - b^2 + 2abi & = & 0 + 1i \\ 2ab & = & 1 & \mbox{looking at the imaginary component} \\ \implies b & = & \frac{1}{2a} \\ a^2 - b^2 & = & 0 & \mbox{looking at the real component} \\ a^2 - (\frac{1}{2a})^2 & = & 0 \\ a^4 - \frac{1}{4} & = & 0 \\ a & = & \pm \frac{1}{\sqrt{2}} \\ b & = & \frac{1}{2a} \\ & = & \pm \frac{1}{\sqrt{2}} \\ \implies w & = & \pm \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right)\end{eqnarray}$$

(noting that the equation for $b$ means that $a$ and $b$ must always have the same sign, which is why there are only two solutions).