Question

Solve:

$\sqrt{4+3\sqrt{20}i}$

• A. $\sqrt{5}+2i$
• B. $3+\sqrt{5}i$
• C. $1+\sqrt{5}i$
• D. $\sqrt{5i}-2i$

Answer 1

The correct option is B $3+\sqrt{5}i$

$Letx+iy=\sqrt{4+3\sqrt{20}i}$

$\therefore (x+iy{)}^2=4+3\sqrt{20}i$

expanding ,

$x^2+2ixy-y^2=4+3\sqrt{20}i$

equating real & imaginary parts

$x^2-y^2=4-----------\left(1\right)$

$2xy=3\sqrt{20}----------------\left(2\right)$

By equaling the modules

$\left|\right(x+iy{)}^2|=|4+3\sqrt{20}i|$

$\therefore x^2+y^2=\sqrt{4^2+(3\sqrt{20}{)}^2}=\sqrt{16+180}=\sqrt{196}=14$

$\therefore x^2+y^2=14----------\left(3\right)$

from (1)&(3),

$x=\pm 3$

$y=\pm \sqrt{5}$ - $(\because e{q}^n$ (2) demands x & y to have same sign)

$\therefore ans:\pm (3+\sqrt{5}i)$