Question

Express the following in the form a +bi, where a and b are real numbers.

Answer 1

$\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{(1+i{)}^2}{1^2-i^2}=\frac{1+2i+i^2}{1+1}=\frac{1+2i-1}{2}=i=0+1i.\left(ii\right){\left(\frac{1-i}{1+i}\right)}^2=\frac{1-2i+i^2}{1+2i+i^2}=\frac{1-2i-1}{1+2i-1}=\frac{-2i}{2i}=-1=-1+0i.\left(iii\right)\left(\frac{1+i}{1-i}\right)=[\frac{1+i}{1-i}\times \frac{1+i}{1+i}]=\frac{(1+i{)}^2}{1-i^2}=\frac{1+i^2+2i}{1-(-1)}=\frac{1-1+2i}{2}=i\therefore {\left(\frac{1+i}{1-i}\right)}^3=i^3=i.i^2=-i=0-1i.\left[Using\right(i\left)\right]$
$\left(iv\right)\frac{(3+\sqrt{5i})(3-\sqrt{5}i)}{(\sqrt{3}+\sqrt{2i})-(\sqrt{3}-\sqrt{2i})}=\frac{9-5i^2}{2\sqrt{2i}}=\frac{9+5}{2\sqrt{2i}}=\frac{14}{2\sqrt{2i}}=\frac{14}{2\sqrt{2}i}\frac{i}{i}=\frac{14i}{2\sqrt{2}{i}^2}=\frac{7\sqrt{2}}{2}i.$