Question

Express in the form $A+iB$
(1) $\frac{3+5i}{2-3i}$
(2) $\frac{\sqrt{3}-i\sqrt{2}}{2\sqrt{3}-i\sqrt{2}}$
(3) $\frac{1+i}{1-i}$
(4) $\frac{{(1+i)}^2}{3-i}$
(5) $\frac{{(a+ib)}^2}{a-ib}-\frac{{(a-ib)}^2}{a+ib}$

Answer 1

We know that $(a+ib)(a-ib)=a^2+b^2$

Here to express in $A+iB$ form we have to make the denominator real we have to multiply both denominator and numerator with complex conjugate of denominator.

$\left(1\right)$ $\frac{3+5i}{2-3i}$

multiply both denominator and numerator with $2+3i$

$=\frac{(3+5i)(2+3i)}{(2+3i)(2-3i)}$

$=\frac{-9+19i}{13}$

$\left(2\right)$ $\frac{\sqrt{3}-i\sqrt{2}}{2\sqrt{3}-i\sqrt{2}}$

multiply both denominator and numerator with $2\sqrt{3}+i\sqrt{2}$

$\frac{(\sqrt{3}-i\sqrt{2})(2\sqrt{3}+i\sqrt{2})}{(2\sqrt{3}-i\sqrt{2})(2\sqrt{3}+i\sqrt{2})}$

$=\frac{8-i\sqrt{6}}{14}$

$\left(3\right)$ $\frac{1+i}{1-i}$

multiply both denominator and numerator with $1+i$

$=\frac{(1+i)(1+i)}{(1+i)(1-i)}$

$=\frac{2i}{2}$

$=i$

$\left(4\right)$ $\frac{(1+i{)}^2}{3-i}$

multiply both denominator and numerator with $3+i$

$=\frac{(1+i{)}^2(3+i)}{(3+i)(3-i)}$

$=\frac{2i(3+i)}{10}$

$=\frac{-1+3i}{5}$

$\left(5\right)$ $\frac{(a+ib{)}^2}{a-ib}-\frac{(a-ib{)}^2}{a+ib}$

cross multiplying and solving

$=\frac{(a+ib{)}^3-(a-ib{)}^3}{a^2+b^2}$

We know that $p^3-q^3=(p-q)(p^2+pq+q^2)$,hence expanding the numerator we get

$=\frac{\left(2ib\right)(3a^2-b^2)}{a^2+b^2}$