JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Complex Numbers

Write $z=\frac{\sqrt{3}}{2}+\frac{i}{2}=e^{i\pi /6}$, which has modulus $1$ and argument $\frac{\pi }{6}$. Then $iz=e^{i\pi /2}{e}^{i\pi /6}=e^{i2\pi /3}$, $z^5=e^{i5\pi /6}$, and $i{z}^8=e^{i\pi /2}{e}^{i4\pi /3}=e^{i11\pi /6}$. Adding $1$ to these three and converting to rectangular form, the real parts give $1-\frac{1}{2}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{1}{2}$ and the imaginary parts give $\frac{\sqrt{3}}{2}+\frac{1}{2}-\frac{1}{2}=\frac{\sqrt{3}}{2}$, so the bracket is $\frac{1}{2}+i\frac{\sqrt{3}}{2}=e^{i\pi /3}$. Raising to the ninth power, ${\left(e^{i\pi /3}\right)}^9=e^{i3\pi }=-1$.