Question

If $z_1=1+\sqrt{3}i$, $z_2=1-\sqrt{3}i,$ then $\frac{z_1^{100}+z_2^{100}}{z_1+z_2}=$

• A. $2^{100}$
• B. -$2^{100}$
• C. $2^{99}$
• D. $-2^{99}$

Answer 1

Answer: (C)

$2^{99}$

$z_1$ $=1+\sqrt{3}i=$$2w$

$z_2$ $=1-\sqrt{3}i=$$2w^2$

We know , $w^2+w=-1$ and $w^3=1$

$=\frac{{z_1}^{100}+{z_2}^{100}}{z_1+z_2}$

$=\frac{(2w{)}^{100}+{\left(2w^2\right)}^{100}}{2w+2w^2}$

$=\frac{2^{100}\left(\right(w{)}^{100}+{\left(w^2\right)}^{100})}{2(w+w^2)}$

$=\frac{2^{100}\left(\right(w)+\left(w^2\right))}{2(-1)}$

$=\frac{2^{100}(-1)}{2(-1)}$

$=$ $2^{99}$