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If $P$ is a complex number whose modulus is one, then the equation $\left(\frac{1+i z}{1-i z}\right)^4=P$ has
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Verified Answer
The correct answer is:
all complex roots
We have,
$$
\begin{array}{rlrl}
& & \left(\frac{1+i z}{1-i z}\right)^4 & =P \\
\Rightarrow & \left(\frac{i-z}{i+z}\right)^4 & =P \\
\Rightarrow & \left(\frac{z-i}{z+i}\right)^4 & =P \\
\Rightarrow & & \left|\frac{z-i}{z+i}\right|^4 & =|P| \\
\Rightarrow & & \left|\frac{z-i}{z+i}\right|^4 & =1 \\
\Rightarrow & & |z-i|^4 & =|z+i|^4 \\
\Rightarrow & & |z-i| & =|z+i|
\end{array}
$$
$\therefore z$ lies on perpendicular bisector of $i$ and $-i$.
$\therefore z$ lies on $y$-axis.
$\therefore z$ has all complex roots.
$$
\begin{array}{rlrl}
& & \left(\frac{1+i z}{1-i z}\right)^4 & =P \\
\Rightarrow & \left(\frac{i-z}{i+z}\right)^4 & =P \\
\Rightarrow & \left(\frac{z-i}{z+i}\right)^4 & =P \\
\Rightarrow & & \left|\frac{z-i}{z+i}\right|^4 & =|P| \\
\Rightarrow & & \left|\frac{z-i}{z+i}\right|^4 & =1 \\
\Rightarrow & & |z-i|^4 & =|z+i|^4 \\
\Rightarrow & & |z-i| & =|z+i|
\end{array}
$$
$\therefore z$ lies on perpendicular bisector of $i$ and $-i$.
$\therefore z$ lies on $y$-axis.
$\therefore z$ has all complex roots.
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