Question

The condition for the cube of $a+ib$ to be a real number is

• A. $a=0$ or $a=\pm \sqrt{3}b$
• B. $a=0$ or $b=\pm \sqrt{3}b$
• C. $b=0$ or $b=\pm \sqrt{3}a$
• D. $b=0$ or $a=\pm \sqrt{3}a$

Answer 1

The correct option is C $b=0$ or $b=\pm \sqrt{3}a$

$z=(a+ib)$
Cubing both sides
$z^3=(a+ib{)}^3$
$=a^3+(ib{)}^3+i3a^{2}b-3a{b}^2$
$=a^3-3a{b}^2+i(3a^{2}b-b^3)$
Since $z^3$ is purely real, hence
$3a^{2}b-b^3=0$
Or
$b(3a^2-b^2)=0$
$b=0$ or $b^2=3a^2$
$b=\pm \sqrt{3}a$.
Hence
$b=0,\pm \sqrt{3}a$