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The solutions of the equation $z^2\left(1-z^2\right)=16$, $z \in \mathbf{C}$, lie on the curve
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Verified Answer
The correct answer is:
$|z|=2$
Given equation, $z^2\left(1-z^2\right)=16, z \in \mathbf{C}$
Now, let $z^2=w=r(\cos \theta+i \sin \theta)$ where $r>0$.
$\begin{array}{lc}
\therefore & 1-z^2=\frac{16}{z^2} \Rightarrow z^2+\frac{16}{z^2}=1 \\
\Rightarrow & \left(r+\frac{16}{r}\right) \cos \theta+i\left(r-\frac{16}{r}\right) \sin \theta=1
\end{array}$
On comparing real and imaginary parts, we get $\left(r+\frac{16}{r}\right) \cos \theta=1$ and $\left(r-\frac{16}{r}\right) \sin \theta=0$
$\therefore$ Either $\sin \theta=0$
$\begin{aligned}
& \Rightarrow \quad \cos \theta=1 \Rightarrow r+\frac{16}{r}=1 \quad \text { (not possible) } \\
& \text { or } \quad r-\frac{16}{r}=0 \Rightarrow r^2=16 \Rightarrow r=4 \\
& \Rightarrow \quad|z|^2=4 \Rightarrow|z|=2 \\
&
\end{aligned}$
Now, let $z^2=w=r(\cos \theta+i \sin \theta)$ where $r>0$.
$\begin{array}{lc}
\therefore & 1-z^2=\frac{16}{z^2} \Rightarrow z^2+\frac{16}{z^2}=1 \\
\Rightarrow & \left(r+\frac{16}{r}\right) \cos \theta+i\left(r-\frac{16}{r}\right) \sin \theta=1
\end{array}$
On comparing real and imaginary parts, we get $\left(r+\frac{16}{r}\right) \cos \theta=1$ and $\left(r-\frac{16}{r}\right) \sin \theta=0$
$\therefore$ Either $\sin \theta=0$
$\begin{aligned}
& \Rightarrow \quad \cos \theta=1 \Rightarrow r+\frac{16}{r}=1 \quad \text { (not possible) } \\
& \text { or } \quad r-\frac{16}{r}=0 \Rightarrow r^2=16 \Rightarrow r=4 \\
& \Rightarrow \quad|z|^2=4 \Rightarrow|z|=2 \\
&
\end{aligned}$
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