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Solve the binomial equation $$z^4 = -8$$

Below is the steps i have done

1: I have taken |-8| that is 8 and then done 8^(1/4) which is 2^(1/4).

2: Since $z=r(cos\alpha+isin\alpha)$ leads me to

$r^4(cos4\alpha+isin4\alpha)=-8(cos\pi/2+isin\pi/2)$

Divide by 4 since the z term is raised by four gives $2^{1/4}(cos\pi/8 + k * \pi/2 +sin\pi/8 + k * \pi/2) $

Is this the correct way to solve this problem ? I am asking since i just started with binomic equations and been stuck for some hours with the question.

addde
  • 449

3 Answers3

1

Another approach that is an efficient way forward is to exploit Euler's Identity and write

$$z^4=r^4e^{i4\theta}=-8=8e^{i(2n+1)\pi}$$

for all integer $n$. Thus, upon inverting we have for $z$

$$z=re^{i\theta}=2^{3/4}e^{i(2n+1)\pi/4}$$

for $n=\pm 1, \pm 3$. Therefore, the $4$ roots of $z$ are

$$\bbox[5px,border:2px solid #C0A000]{z=2^{3/4}e^{\pm i\pi/4}\,\,\text{and}\,\,2^{3/4}e^{\pm i3\pi/4}}$$

or in rectangular form

$$\bbox[5px,border:2px solid #C0A000]{z=2^{1/4}(1\pm i)\,\,\text{and}\,\,2^{1/4}(-1\pm i)}$$

Mark Viola
  • 179,405
0

$$ z^4 = -8 \Longleftrightarrow $$ $$ z^4 = |-8|e^{\arg(-8)i} \Longleftrightarrow $$ $$ z^4 = 8e^{\pi i} \Longleftrightarrow $$ $$ z^4 = 8e^{\left( \pi + 2 \pi k \right) i} \Longleftrightarrow $$ $$ z = \left( 8e^{\left( \pi + 2 \pi k \right) i} \right)^{\frac{1}{4}} \Longleftrightarrow $$ $$ z = \sqrt[4]{8}e^{\frac{1}{4}\left( \pi + 2 \pi k \right) i} \Longleftrightarrow $$ $$ z = \sqrt[4]{8}e^{\left( \frac{\pi}{4} + \frac{\pi k}{2} \right) i} \Longleftrightarrow $$ $$ z = \sqrt[4]{8}e^{\frac{\pi(2k+1)}{4} i} $$

With $k \in \mathbb{Z}$ and $k$ goes from $0-3$

So the solutions are:

$$z_0=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 0+1)}{4} i}=\sqrt[4]{8}e^{\frac{\pi}{4}i}$$ $$z_1=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 1+1)}{4} i}=\sqrt[4]{8}e^{\frac{3\pi}{4}i}$$ $$z_2=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 2+1)}{4} i}=\sqrt[4]{8}e^{-\frac{3\pi}{4}i}$$ $$z_3=\sqrt[4]{8}e^{\frac{\pi(2 \cdot 3+1)}{4} i}=\sqrt[4]{8}e^{-\frac{\pi}{4}i}$$

Jan Eerland
  • 28,671
-1

Notice, $$z^4=-8$$ $$z^4=8i^2$$ $$z=\sqrt[4]{8i^2}$$ $$z=(2)^{3/4}\sqrt{i}$$ $$z=(2)^{3/4}\sqrt{\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}}$$ $$=(2)^{3/4}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{1/2}$$ $$=(2)^{3/4}\left(\cos\left(2k\pi+\frac{\pi}{2}\right)+i\sin\left(2k\pi+\frac{\pi}{2}\right)\right)^{1/2}$$ $$=(2)^{3/4}\left(\cos\left(\frac{(4k+1)\pi}{2}\right)+i\sin\left(\frac{(4k+1)\pi}{2}\right)\right)^{1/2}$$ $$=(2)^{3/4}\left(\cos\left(\frac{(4k+1)\pi}{4}\right)+i\sin\left(\frac{(4k+1)\pi}{4}\right)\right)$$ Setting $k=0$, we get $$z=(2)^{3/4}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)$$ $$=(2)^{3/4}\left(\frac{1}{\sqrt 2}+i\frac{1}{\sqrt 2}\right)$$ $$z=(2)^{1/4}\left(1+i\right)$$

Setting $k=1$, we get $$z=(2)^{3/4}\left(\cos\left(\frac{5\pi}{4}\right)+i\sin\left(\frac{5\pi}{4}\right)\right)$$ $$=(2)^{3/4}\left(\frac{-1}{\sqrt 2}+i\frac{-1}{\sqrt 2}\right)$$ $$z=-(2)^{1/4}\left(1+i\right)$$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=(2)^{1/4}\left(1+i\right),\ z=-(2)^{1/4}\left(1+i\right)}}$$

Harish Chandra Rajpoot
  • 37,450