Solve $z^4+1=0$ algebraically

Question

I know the result and how to solve it using trigonometry and De Moivre.

However, given that the complex number $z$ can be rewritten as $a+bi$, how can I solve it algebraically?

Answer 1

Wouldn't you just write $$ z^4+1=0\\ \Updownarrow\\ w^2=-1,z^2=w $$ And conclude that $z^2=w=\pm i$. Then with $z=a+bi$ the equation $z^2=i$ becomes $$ (a+bi)^2=i\\ \Updownarrow\\ a^2-b^2+2abi=i $$ leads to $a^2-b^2=0$ and thus $a=\pm b$ and then $2abi=i$ gives us $ab=\frac{1}{2}$. So either $a=b=\pm\frac{\sqrt{2}}{2}$ or $a=-b$ which then yields non-real $a,b$ which is absurd.

Then you can repeat the strategy from above for the case $z^2=-i$ to get the other two solutions.

Answer 2

Inserting $z = a + bi$ we get $$ a^4 - 6a^2b^2 + b^4 + 1 + ab(a - b)(a+b)i = 0 $$ which becomes two real equations $$ a^4 - 6a^2b^2 + b^4 + 1 = 0 \quad \wedge\quad ab(a-b)(a+b) = 0 $$ The second one is already factored for you, giving you four specific cases to work with, two without real solutions. The first one can be seen as a quadratic equation in $a^2$ or $b^2$, depending on which case you choose from the second equation.

Answer 3

Using polar form is the standard method, but there is another trick. Some creative rewriting gives:

$$z^4 + 1 = (z^2+1)^2 - 2z^2 = (z^2 + z\sqrt 2 + 1)(z^2 - z\sqrt 2 +1)$$

and you can reduce the problem to solving two quadratic equations.

Answer 4

$$ z^4 + 1 = 0 \Rightarrow z^4 + 2z^2 + 1 - 2z^2 = 0 $$

$$ \Rightarrow (z^2 + 1)^2 -(\sqrt{2}z)^2 = 0 $$

$$ \Rightarrow (z^2 - \sqrt{2} z + 1)( z^2 + \sqrt{2}z + 1 ) = 0 $$

use $$ ax^2 + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Answer 5

Hints:

• $z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt2z+1)(z^2-\sqrt2z+1)$
• $z^2\pm\sqrt2z+1=(z\pm\tfrac12\sqrt2)^2+\tfrac12$
• $u^2+a^2=(u+\mathrm ia)(u-\mathrm ia)$

Answer 6

Hint: $z^4+1=(z^2+i)(z^2-i)$ and $i=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)^2$ and $-i=(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)^2$ therefore: $$ z^4+1=\left(z-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)\right)\left(z+(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)\right)\left(z-(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\right)\left(z+(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\right) $$

Answer 7

With total cheat, use the change of variable $w=\dfrac{1+i}{\sqrt2}z$.

The equation becomes

$$w^4-1=0$$

with the obvious solutions $$w=\pm1,\pm i.$$

Answer 8

1. Question Summary With Adjusted Formatting

To make it easier to follow the answer, the question is summarized below:

$$ \text{Let }Z^4+1=0, Z=a+b i \tag{Eq. 1.1} $$

I know the result and how to solve it using trigonometry and De Moivre.

However, given that the complex number z can be rewritten as a+bi, how can I solve it algebraically?

2. Algebraic Steps to Check Results (Verified Correct)

Algebraically, the following table can be used to determine the algebraic coefficients: $$ 1\\ 1\text{ }2\text{ }1\\ 1\text{ }3\text{ }3\text{ }1\\ 1\text{ }4\text{ }6\text{ }4\text{ }1\\ \tag{Table 2.1}$$ Then,$$Z^4=\left(a + b i \right)^4 =\\ =a^4 + 4a^3 (b i) + 6a^2 (b i)^2 + 4a (b i)^3 + (b i)^4 \tag{Eq. 2.1}$$ Using the following identities: $$(b i)^2=-b^2,(b i)^3=-i b^3, (b i)^4=b^4$$ It follows that: $$Z^4=a^4 + 4 a^3 (b i) - 6 a^2 b^2 + 4a (-i b^3) + b^4 \\ \left( a^4 - 6 a^2 b^2 + b^4 \right) + i \left( 4 a^3 (b) - 4a (b^3) \right) =-1 \tag{Eq. 2.2}$$

Separating real and imaginary quantities yields: $$ \left( a^4 - 6 a^2 b^2 + b^4 \right) = -1 \tag{Eq. 2.3} $$ $$ i \left( 4 a^3 (b) - 4a (b^3) \right) = 0 \tag{Eq. 2.4} $$ From Equation 2.4 (dividing by $4 i a b $): $$ a^2 - b^2 = 0 \longrightarrow a= ±b \tag{Eq. 2.5} $$ From checking the trigonometric solutions for $a$ and $b$, in Equations 3.3a, 3.3b, 3.3c, and 3.3d, the algebraic result of Equation 2.5 is correct. It is the same result using trigonometry as is yielded algebraically.

Further substituting this $b=±a$ into Equation 2.3 yields: $$ \left( a^4 - 6 a^2 (±a)^2 + (±a)^4 \right) = -1 \\ \longrightarrow a^4=\frac{1}{4}=\frac{1}{2^2} \tag{Eq. 2.5} $$

Therefore, $$ a=±\frac{1}{\sqrt{2}} \text{ and } b=±a \tag{Eq. 2.6} $$

yielding algebraically exactly the same result as with trigonometry. Formally the four solutions from Equation 2.5 are: $$ a=\frac{1}{\sqrt{2}}, b=a=\frac{1}{\sqrt{2}} \\ a=\frac{1}{\sqrt{2}}, b=-a=-\frac{1}{\sqrt{2}} \\ a=-\frac{1}{\sqrt{2}}, b=a=-\frac{1}{\sqrt{2}} \\ a=-\frac{1}{\sqrt{2}}, b=-a=\frac{1}{\sqrt{2}} \\ \tag{Eqations 2.7a, 2.7b, 2.7c, 2.7d} $$

The results have been verified against the trigonometric results, following the initial flow of the math from the initial submission, but correcting all the errors so that everything now is completely consistent, and the algebraic result is exactly the same as the trigonometric result. For comparison convenience, the trigonometric result is also derived below.

Section 3. Checking the Resulting Answer

From Euler's identity, $$ e^{i \theta}=cos(\theta)+i sin(\theta) \tag{3.1} $$ The equation under consideration $Z^4+1=0$ is a fourth order polynomial and it has four (possibly complex) solutions. $$ -1 = e^{(+ i \pi)}, \\ -1 = e^{(- i \pi)}, \\ -1 = e^{(+i 3 \pi)}, \\ -1 = e^{(-i 3 \pi)} \tag{3.2 a, 3.2 b, 3.2 c, 3.2 d}$$

Then by taking the $4$ roots of $Z^4=-1$ above, the answers are:

$$ Z_a = e^{(+i \frac{1}{4} \pi)} = \frac{1}{\sqrt{2}} \left(1 + i \right) , \\ Z_b = e^{(-i \frac{1}{4} \pi)} = \frac{1}{\sqrt{2}} \left(1 - i \right) , \\ Z_c = e^{(+i \frac{3}{4} \pi)} = \frac{1}{\sqrt{2}} \left(-1 + i \right) , \\ Z_d = e^{(-i \frac{3}{4} \pi)} = \frac{1}{\sqrt{2}} \left(-1 - i \right) \tag{3.3a, 3.3b, 3.3c, 3.3d}$$

These are the so-called trigonometric solutions for $Z$. The algebraic solutions for $Z$ can be tested step-by-step using these solutions.