Chapter 9, Problem 50RE

To determine

To Find: All the complex fourth root of $\left(-16\right)$ .

Answer to Problem 50RE

The complex fourth root of $\left(-16\right)$ is $2i^{\frac{1}{2}},2i^{\frac{3}{2}},-2i^{\frac{1}{2}},-2i^{\frac{3}{2}}$ .

Explanation of Solution

Given information:

The complex fourth root of $\left(-16\right)$ .

Formula:

For complex fourth root $z^{\frac{1}{n}}=r^{\frac{1}{n}}\cos \left(\frac{\theta }{n}+\frac{K{.360}^0}{n}\right)$ where K is from $1,2,\mathrm{.....}\left(n-1\right)$ .

Calculation:

Consider the statement the complex fourth root of $\left(-16\right)$ .

Recall that for the fourth root of unity of $\left(-16\right)$ then $z^{\frac{1}{n}}=r^{\frac{1}{n}}\cos \left(\frac{\theta }{n}+\frac{K{.360}^0}{n}\right)$ .

Therefore $z=-16$

  $\Rightarrow z^4=-16\Rightarrow z={\left(-16\right)}^{\frac{1}{4}}$ where $n=4$ .

Therefore $z^{\frac{1}{4}}={\left(-16\right)}^{\frac{1}{4}}\cos \left(\frac{0^0}{4}+K*{90}^0\right)\Rightarrow {\left(-1\right)}^{\frac{1}{4}}{\left(16\right)}^{\frac{1}{4}}\cos \left(\frac{0^0}{4}+K*{90}^0\right)\Rightarrow {\left(i^2\right)}^{\frac{1}{4}}{\left(2^4\right)}^{\frac{1}{4}}\cos \left(\frac{0^0}{4}+K*{90}^0\right)\Rightarrow {\left(i\right)}^{\frac{1}{2}}\left(2\right)\cos \left(\frac{0^0}{4}+K*{90}^0\right)$

For $K=0$

  $z_1=2i^{\frac{1}{2}}\cos 0^0\Rightarrow 2i^{\frac{1}{2}}$ .

For $K=1$

  $z_2=2i^{\frac{1}{2}}\cos {90}^0\Rightarrow 2i^{\frac{1}{2}}*i\Rightarrow 2i^{\frac{3}{2}}$ .

For $K=2$

  $z_3=2i^{\frac{1}{2}}\cos {180}^0\Rightarrow -2i^{\frac{1}{2}}$ .

For $K=3$

  $z_4=2i^{\frac{1}{2}}\cos {270}^0\Rightarrow -2i^{\frac{3}{2}}$ .

Therefore all the complex fourth root of unity of $\left(-16\right)$ is $2i^{\frac{1}{2}},2i^{\frac{3}{2}},-2i^{\frac{1}{2}},-2i^{\frac{3}{2}}$ .