Question

If $\frac{1+2i}{2+i}=r(\cos \theta +i\sin \theta )$ then $r$ and $\theta$ equal

Answer 1

Step 1: Rationalize the left hand side of given equation:

The given equation is $\frac{1+2i}{2+i}=r(\cos \theta +i\sin \theta )$

$\begin{array}{ll}\frac{1+2i}{2+i}& =r(\cos \theta +i\sin \theta )\end{array}$

$\Rightarrow$$\begin{array}{ll}\frac{(1+2i)(2-i)}{(2+i)(2-i)}& =r(\cos \theta +i\sin \theta )\end{array}$

$\Rightarrow$ $\begin{array}{ll}\frac{2+4i-i+2}{4+1}& =r(\cos \theta +i\sin \theta )\end{array}$

$\Rightarrow$ $\begin{array}{ll}\frac{4+3i}{5}& =r(\cos \theta +i\sin \theta )\end{array}$

Step 2: Find the value of $r\cos \left(\theta \right)$ and $r\sin \left(\theta \right)$

$r\cos \left(\theta \right)=\frac{4}{5}$….(1)

$r\sin \left(\theta \right)=\frac{3}{5}$ ….(2)

Step 3: Squaring and adding equation $\left(1\right)$ and equation $\left(2\right)$ and calculate the value of $r$ and $\theta$.

$\begin{array}{rcl}{r}^2[{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)]& =& \frac{16+9}{25}\\ & =& \frac{25}{25}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{r}^2& =& 1\end{array}$

$\Rightarrow$ $\begin{array}{rcl}r& =& 1\end{array}$

$\begin{array}{rcl}\tan \left(\theta \right)& =& \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\\ & =& \frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}}\\ & =& \frac{3}{4}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\theta & =& {\tan }^{-1}\frac{3}{4}\end{array}$

Hence, the values are $\mathit{r}\mathbf{=}\mathbf{1}$ and $\mathit{\theta }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\frac{\mathbf{3}}{\mathbf{4}}$