Question

Let $z=9+bi$ where b is non zero real and $i^2=-1$. If the imaginary part of $z^2$ and $z^3$ are equal, then $b^2$ equals.

• A. $261$
• B. $225$
• C. $125$
• D. $361$

Answer 1

The correct option is B $225$

Given,

$z=9+bi$

Where $b$ is non zero real

$i^2=-1$

If the imaginary part of $z^2$ and$z^3$ are equal,

To Find value of $b^2$

$z^2=(9+bi)(9+bi)$

$=81+b^2{i}^2+18bi$

$=81-b^2+18bi$

imaginary part of $z^2=18b⟶\left(1\right)$

$z^3=\left(z^2\right)\left(z\right)=(81-b^2+18bi)(9+bi)$

$z^3=729+81bi-9b^2+b^{3}i+162bi+18b^2{i}^2$

$z^3=729+243bi-9b^2-b^{3}i-18b^2$

imagenary part of$z^3=243b-b^3⟶\left(2\right)$

given imaginary part of$z^2=$ imagenary part of $z^3$

$18b=243b-b^3$

$b^3=225bb(b^2-225)=0$

$b^2=225$

Hence, the value of $b^2$ is $225.$