Question

The value of 3 minus i root 2 whole square by 1 + 2i

Answer 1

$\displaystyle\mathsf{\therefore\frac{(3-i\sqrt{2})^{2}}{1+2i}}$

$\displaystyle\mathsf{=\frac{9-6\sqrt{2}\:i+2i^{2}}{1+2i}}$

$\displaystyle\mathsf{=\frac{9-6\sqrt{2}\:i-2}{1+2i}}$

$\displaystyle\mathsf{=\frac{7-6\sqrt{2}\:i}{1+2i}}$

$\displaystyle\mathsf{=\frac{(7-6\sqrt{2}\:i)(1-2i)}{(1+2i)(1-2i)}}$

$\displaystyle\mathsf{=\frac{7-14i-6\sqrt{2}\:i+12\sqrt{2}\:i^{2}}{1-4i^{2}}}$

$\displaystyle\mathsf{=\frac{7-14i-6\sqrt{2}\:i-12\sqrt{2}}{1+4}}$

$\displaystyle\mathsf{=\frac{1}{5}\:\{(7-12\sqrt{2})-(14+6\sqrt{2})i\}}$