Chapter 6, Problem 22CT

To determine

To Simplify:

  $\frac{6+i}{3-2i}$

Answer to Problem 22CT

  $\frac{16+15i}{13}$

Explanation of Solution

Given:

Given expression is :

  $\frac{6+i}{3-2i}$

Calculation:

  $\begin{array}{l}\frac{6+i}{3-2i}\\ =\frac{6+i}{3-2i}\cdot \frac{3+2i}{3+2i}\mathrm{.................}[\text{rationalize the denominator]}\\ \text{=}\frac{\left(6+i\right)\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}\\ =\frac{\left(6+i\right)\left(3+2i\right)}{{\left(3\right)}^2-{\left(2i\right)}^2}\mathrm{...............}[(a+b)(a-b)=a^2-b^2]\\ =\frac{6\left(3+2i\right)+i\left(3+2i\right)}{9+4}\mathrm{........}[i^2=-1]\\ =\frac{18+12i+3i-2}{13}\\ =\frac{16+15i}{13}\end{array}$