Browse Questions

If $\;z_{1} = \sqrt{3} + i \sqrt{3}\;$ and $\;z_{2} =\sqrt{3} + i\;$ , then find the quadrant in which $\;(\large\frac{z_{1}}{z_{2}})\;$ lies

$(a)\;(\large\frac{3+ \sqrt{3}}{4})+(\large\frac{3- \sqrt{3}}{4})i\qquad(b)\;(\large\frac{5+ \sqrt{3}}{4})+(\large\frac{5- \sqrt{3}}{4})i\qquad(c)\;(\large\frac{3+ \sqrt{5}}{4})+(\large\frac{3- \sqrt{5}}{4})i\qquad(d)\;(\large\frac{7+ \sqrt{3}}{4})+(\large\frac{7- \sqrt{3}}{4})i$

Answer : $\;(\large\frac{3+ \sqrt{3}}{4})+(\large\frac{3- \sqrt{3}}{4})i$
Explanation :
$\large\frac{z_{1}}{z_{2}} = \large\frac{ \sqrt{3} + i \sqrt{3}}{\sqrt{3} +i}$
$=(\large\frac{3+ \sqrt{3}}{4})+(\large\frac{3- \sqrt{3}}{4})i$
Which is represented by a point in first quadrant .