Question

If $3^{50}(x+iy)={(\frac{3}{2}+\frac{i\sqrt{3}}{2})}^{100}\forall x,yϵR$,then ordered pair (x,y) is given by

• A. $(-\frac{1}{2},\frac{\sqrt{3}i}{2})$
• B. $(\frac{1}{2},\frac{\sqrt{3}i}{2})$
• C. $(-\frac{1}{2},-\frac{\sqrt{3}i}{2})$
• D. $(-\frac{3}{2},-\frac{\sqrt{1}i}{2})$

Answer 1

The correct option is A $(-\frac{1}{2},\frac{\sqrt{3}i}{2})$

The above expression can be simplified as
$\left(i\sqrt{3}{)}^{100}\right(\frac{1}{2}-\frac{\sqrt{3}i}{2}{)}^{100}$
$=3^{50}(-w{)}^{100}$
$=3^{50}(w{)}^{100}$
$=3^{50}(w{)}^{99}.w$
$=3^{50}(w^3{)}^{33}.w$
$=3^{50}w$
Now $w=\frac{-1}{2}+\frac{\sqrt{3}i}{2}$
$=x+iy$
Hence $x=\frac{-1}{2}$ and $y=\frac{\sqrt{3}i}{2}$.