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If $z \in C$ and $i z^3+4 z^2-z+4 i=0$, then a complex root of this equation having minimum magnitude is
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2811 Upvotes
Verified Answer
The correct answer is:
$\frac{1-i}{\sqrt{2}}$
Given complex equation,
$$
\begin{array}{cc}
& i z^3+4 z^2-z+4 i=0 \\
\Rightarrow & z^2(i z+4)+i(i z+4)=0 \\
\Rightarrow & (i z+4)\left(z^2+i\right)=0 \\
\Rightarrow & z=4 i \text { or } z^2=i \\
\Rightarrow & z=4 i \\
\text { or } & \pm\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i\right)
\end{array}
$$
So, the complex root having minimum magnitude is
$$
\frac{1-i}{\sqrt{2}} \text { or } \frac{-1+i}{\sqrt{2}}
$$
$$
\begin{array}{cc}
& i z^3+4 z^2-z+4 i=0 \\
\Rightarrow & z^2(i z+4)+i(i z+4)=0 \\
\Rightarrow & (i z+4)\left(z^2+i\right)=0 \\
\Rightarrow & z=4 i \text { or } z^2=i \\
\Rightarrow & z=4 i \\
\text { or } & \pm\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i\right)
\end{array}
$$
So, the complex root having minimum magnitude is
$$
\frac{1-i}{\sqrt{2}} \text { or } \frac{-1+i}{\sqrt{2}}
$$
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