Chapter 9, Problem 33RE

To determine

To write: Complex number $0.1\left(\cos {350}^0+i\sin {350}^0\right)$ in the standard form of $\left(a+ib\right)$ .

Answer to Problem 33RE

The mentioned complex number in the standard form is $0.0984-i0.0173$ .

Explanation of Solution

Given information:

The mentioned complex number is $0.1\left(\cos {350}^0+i\sin {350}^0\right)$ .

Formula used:

By the use of De Moivre’s theorem we know that if $z=r\left(\cos \theta +i\sin \theta \right)$ then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ where n is any integral value.

Also trigonometric formula $\cos \left({360}^0-\theta \right)=\cos \theta$ and $\sin \left({360}^0-\theta \right)=-\sin \theta$ .

Calculation:

Consider the mentioned complex number is $0.1\left(\cos {350}^0+i\sin {350}^0\right)$ .

Recall the De Moivre’s theorem that if $z=r\left(\cos \theta +i\sin \theta \right)$ then $z^n=r^n\left(\cos n\theta +i\sin n\theta \right)$ where n is any integral value.

Therefore the mentioned complex number is $0.1\left(\cos {350}^0+i\sin {350}^0\right)$ which is in the form of $z=r\left(\cos \theta +i\sin \theta \right)$ .

Here $r=0.1$ , $\theta$ is the angle which is equal to ${350}^0$ and $n=1$ .

Also recall the trigonometric formula $\cos \left({360}^0-\theta \right)=\cos \theta$ and $\sin \left({360}^0-\theta \right)=-\sin \theta$ .

Therefore by De Moivre’s theorem we have,

  $\begin{array}{l}z=0.1\left(\cos {350}^0+i\sin {350}^0\right)\Rightarrow z=0.1\left(\cos \left({360}^0-{10}^0\right)+i\sin \left({360}^0-{10}^0\right)\right)\Rightarrow z=\frac{1}{10}\left(\cos {10}^0-i\sin {10}^0\right)\Rightarrow z=\left(\frac{\cos {10}^0}{10}-i\frac{\sin {10}^0}{10}\right)\\ \Rightarrow z=\left(\frac{0.984}{10}-i\frac{0.173}{10}\right)\Rightarrow z=\left(0.0984-i0.0173\right)\end{array}$

Thus the required Complex number $0.1\left(\cos {350}^0+i\sin {350}^0\right)$ in the standard form of $\left(a+bi\right)$ is $0.0984-i0.0173$ .