Question

If $z=(\frac{\sqrt{3}}{2}+\frac{i}{2}{)}^5+(\frac{\sqrt{3}}{2}-\frac{i}{2}{)}^5$, then

• A. $Re\left(z\right)=0$
• B. $Im\left(z\right)=0$
• C. $Re\left(z\right)>0$, $Im\left(z\right)>0$
• D. $Re\left(z\right)>0$,$Im\left(z\right)<0$

Answer 1

The correct option is B $Im\left(z\right)=0$

Given,

$z={(\frac{\sqrt{3}}{2}+\frac{i}{2})}^5+{(\frac{\sqrt{3}}{2}-\frac{i}{2})}^5$

$=\frac{{(\sqrt{3}+i)}^5}{2^5}+\frac{{(\sqrt{3}-i)}^5}{2^5}$

$=\frac{{(\sqrt{3}+i)}^5+{(\sqrt{3}-i)}^5}{2^5}$

$=\frac{\sqrt{3}(16+{(\sqrt{3}+i)}^4+{(\sqrt{3}-i)}^4-(8+8\sqrt{3}i)-(8-8\sqrt{3}i))}{2^4}$

$=\frac{\sqrt{3}(16+{(\sqrt{3}+i)}^4+{(\sqrt{3}-i)}^4-(8+8\sqrt{3}i)-(8-8\sqrt{3}i))}{16}$

$=\frac{(-16)\sqrt{3}}{16}$

$=-\sqrt{3}+i\times 0$

$z=x+iy$

real part of $Re\left(z\right)=-\sqrt{3}$

imaginary part of $Im\left(z\right)=0$