Answer:
(a) \frac{-1i}{2}-1[/tex]
(b) [tex]\frac{20i+21}{29}[/tex]
(c) i
Step-by-step explanation:
We have to perform division
(a) [tex]\frac{1-2i}{2i}[/tex]
So after division
[tex]\frac{1-2i}{2i}=\frac{1}{2i}-\frac{2i}{2i}=\frac{-1i}{2}-1[/tex]
(b) We have given expression [tex]\frac{5-2i}{5+2i}[/tex]
After rationalizing [tex]\frac{5-2i}{5+2i}\times \frac{5-2i}{5-2i}=\frac{(5-2i)^2}{25+4}=\frac{25+4i^2+20i}{29}=\frac{20i+21}{29}[/tex]
(c) We have given expression [tex]\frac{\sqrt{3}-2i}{-2-\sqrt{3i}}[/tex]
After rationalizing
[tex]\frac{\sqrt{3}-2i}{-2-\sqrt{3i}}\times \frac{-2+\sqrt{3i}}{-2+\sqrt{3i}}=\frac{-2\sqrt{3}+7i+2\sqrt{3}}{7}=\frac{7i}{7}=i[/tex]
