Question

Let $z=9+bi$, where $b$ is nonzero real and $i^2=-1$. If the imaginary part of $z^2$ and $z^3$ are equal, then find the value of $\frac{b}{3}$.

Answer 1

Given that, $z=9+bi$

Since, imaginary parts of $z^2$ & $z^3$ are equal

Therefore, $\frac{z^2-{\overline{z}}^2}{2i}=\frac{z^3-{\overline{z}}^3}{2i}$

$\Rightarrow (z-\overline{z})(z+\overline{z})=(z-\overline{z})(z^2+{\overline{z}}^2+z\overline{z})$

$\Rightarrow z+\overline{z}={(z+\overline{z})}^2-z\overline{z}$

$\Rightarrow z\overline{z}={18}^2-18=306$

$\Rightarrow {\left|z\right|}^2=306$

$\Rightarrow 81+b^2=306$

$\Rightarrow b=15$

$\Rightarrow \frac{b}{3}=5$

Ans: 5