Question

Let $Z$ be a complex number satisfying the equation ${(Z^2+3)}^2=-16$ then $\left|Z\right|$ has the value equal to

• A. $5^{1/2}$
• B. $5^{1/3}$
• C. $5^{2/3}$
• D. $5$

Answer 1

The correct option is A $5^{1/2}$

Given $(z^2+3{)}^2=-16|z|=?$

$x^2+3=\pm 4i$ removing square on both sides

Now we know $z=a+ib$

$\Rightarrow (a+ib{)}^2+3=\pm 4i$

$\Rightarrow a^2-b^2+2abi+3=\pm 4i$

equating real part to real and imaginary to imaginary

$a^2-b^2+3=0$ - $\left(1\right)$

$2ab=\pm 4$ - $\left(2\right)$ $ab=\pm 2b=\pm \frac{2}{a}$ - $\left(3\right)$

put $\left(3\right)$ in eq $\left(1\right)$

$\Rightarrow a^2-{\left(\frac{2}{a}\right)}^2+3=0\Rightarrow a^2-\frac{4}{a^2}+3=0$

$\Rightarrow a^4-1+3a^2=0$

Let $a^2=t$

$\Rightarrow t^2-4+3t=0$

$D=b^2-4ac=9-4\times 1\times -4=9+16$

roots $x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-3\pm 5}{2}$

$x=1$ $y=-4$

so $a^2=1$ or $a^2=-4$

since square cannot be negative this not exist

$a^2=1$ $a=\pm 1$

$b=\frac{\pm 2}{\pm 1}$ $b=\pm 2$

$z=a4+ib=\pm 1+2i$ $|z|=\sqrt{1+4}$ $|z|=\sqrt{5}$