Chapter 6, Problem 25CT

To determine

Thefourth roots of $256\left(1+\sqrt{3}i\right)$ using formula for roots.

Answer to Problem 25CT

Thefourth roots of $256\left(1+\sqrt{3}i\right)$ are $z_0=4.59+1.23i,z_1=-1.23+4.59i,z_2=-4.59-1.23i\text{ and }{z}_3=1.23-4.59i$

Explanation of Solution

Given info:

  $\sqrt[4]{256\left(1+\sqrt{3}i\right)}$

Formula used:

The trigonometric form of the complex number $z=a+ib$ is

  $z=r\left(\cos \theta +i\sin \theta \right)$

Where $r=\sqrt{a^2+b^2}$ and $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

For a positive integer n the complex number $z=r\left(\cos \theta +i\sin \theta \right)$ has exactly n distinct nth roots given by

  $z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)$

where k = 0, 1, 2, . . . , ,n -1.

Calculation:

We have

  $\begin{array}{l}a=256,b=256\sqrt{3}\\ r=\sqrt{a^2+b^2}=\sqrt{{256}^2+{\left(256\sqrt{3}\right)}^2}=512\\ \theta ={\tan }^{-1}\left(\frac{256\sqrt{3}}{256}\right)=\frac{\pi }{3}\end{array}$

  $\begin{array}{l}256\left(1+\sqrt{3}i\right)=512\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\end{array}$

  $\begin{array}{l}n=4,r=512,\theta =\frac{\pi }{3}\\ z_0=\sqrt[4]{512}\left(\cos \left(\frac{\frac{\pi }{3}+2\pi \times 0}{4}\right)+i\sin \left(\frac{\frac{\pi }{3}+2\pi \times 0}{4}\right)\right)\end{array}$

  $\begin{array}{l}{z}_0=4.76\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right)=4.59+1.23i\\ z_1=\sqrt[4]{512}\left(\cos \left(\frac{\frac{\pi }{3}+2\pi \times 1}{4}\right)+i\sin \left(\frac{\frac{\pi }{3}+2\pi \times 1}{4}\right)\right)\end{array}$

  $\begin{array}{l}{z}_1=4.76\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)=-1.23+4.59i\\ z_2=\sqrt[4]{512}\left(\cos \left(\frac{\frac{\pi }{3}+2\pi \times 2}{4}\right)+i\sin \left(\frac{\frac{\pi }{3}+2\pi \times 2}{4}\right)\right)\end{array}$

  $\begin{array}{l}{z}_2=4.76\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)=-4.59-1.23i\\ z_3=\sqrt[4]{512}\left(\cos \left(\frac{\frac{\pi }{3}+2\pi \times 3}{4}\right)+i\sin \left(\frac{\frac{\pi }{3}+2\pi \times 3}{4}\right)\right)\end{array}$

  $z_3=4.76\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)=1.23-4.59i$

Thefourth roots of $256\left(1+\sqrt{3}i\right)$ are $z_0=4.59+1.23i,z_1=-1.23+4.59i,z_2=-4.59-1.23i\text{ and }{z}_3=1.23-4.59i$

Conclusion:

Thus,the fourth roots of $256\left(1+\sqrt{3}i\right)$ are $z_0=4.59+1.23i,z_1=-1.23+4.59i,z_2=-4.59-1.23i\text{ and }{z}_3=1.23-4.59i$