Question

If a = s +ti and b = u+vi are in $\mathbf{\mathbb{Z}}\left|i\right|$and $\mathit{b}\mathbf{\ne }\mathbf{0}$, show that a/b = c+di, where$\mathit{c}\mathbf{=}\frac{\mathbf{s}\mathbf{u}\mathbf{+}\mathbf{t}\mathbf{v}}{{\mathbf{u}}^{\mathbf{2}}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}$ and $\mathit{d}\mathbf{=}\frac{\mathbf{t}\mathbf{u}\mathbf{-}\mathbf{s}\mathbf{v}}{{\mathbf{u}}^{\mathbf{2}}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}$.

Short answer

It has been proved that $a/b=c+di$.

Step-by-step solution

Given that

Given that$a=s+ti$ and$b=u+vi$ are in $\mathrm{\mathbb{Z}}\left|i\right|$and $b\ne 0$.

Find a/b

$a/b=\frac{s+ti}{u+vi}$

Multiplying and dividing by u - vi.

$\begin{array}{rcl}a/b& =& \frac{\left(s+ti\right)\left(u-vi\right)}{\left(u+vi\right)\left(u-vi\right)}\\ & =& \frac{su-svi+uti-vt{i}^2}{u^2-uvi+uvi-v{i}^2}\\ & =& \frac{su-svi+uti+vt}{u^2+v^2}\\ & =& \frac{su+vt}{u^2+v^2}+\frac{\left(ut-sv\right)i}{u^2+v^2}\end{array}$

Here, let $c=\frac{su+tv}{u^2+v^2}$and $d=\frac{tu-sv}{u^2+v^2}$.

Conclusion

Hence, $a/b=c+di$, where $c=\frac{su+tv}{u^2+v^2}$and $d=\frac{tu-sv}{u^2+v^2}$.