Question

If $\mathit{a}\mathbf{=}\mathit{s}\mathbf{+}\mathit{t}\mathit{i}$and $\mathit{b}\mathbf{=}\mathit{u}\mathbf{+}\mathit{v}\mathit{i}$are in $\mathbf{\mathbb{Z}}\left[i\right]$and $\mathit{b}\mathbf{\ne }\mathbf{0}$, show that $\mathit{a}\mathbf{/}\mathit{b}\mathbf{=}\mathit{c}\mathbf{+}\mathit{d}\mathit{i}$, 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  

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

Multiplying and dividing by $u=vi$.

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

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 role="math" localid="1653289634482" $c=\frac{su+tv}{u^2+v^2}$and $d=\frac{tu-sv}{u^2+v^2}$.