College Algebra

$(\frac{1}{5}+\sqrt{\frac{19}{10i}})^2\\ = \frac{1}{25}+2\times \frac{1}{5}\times \sqrt{\frac{19}{10i}}+\frac{19}{10i}\\= \frac{1}{25}+\frac{2}{5}\times\frac{\sqrt{19}}{\sqrt{10}}\times \frac{\sqrt{2}}{1+i}+\frac{19}{10i}$

Because sqrt i = (1+i)/ sqrt2

http://web2.0calc.com/questions/what-is-square-root-of-i

$=\frac{1}{25}+\frac{2\sqrt{38}}{(5\sqrt{10})(1+i)}+\frac{19}{10i}$

$=\frac{1}{25}+\frac{2\sqrt{-380}}{50i(1+i)}+\frac{95+95i}{50i(1+i)}$

$=\frac{1}{25}+\frac{(2\sqrt{380}+95)i+95}{50i-50}$

You can do the last half yourself.