Question

If $4z=1+2i$, then the value of $z^3+7z^2-z+16$

Answer 1

We have,

$4z=1+2i$

$z=\frac{1+2i}{4}$

Since,

$z^3+7z^2-z+16$ …….. (1)

On putting the value of $z$ in equation (1), we get

$={\left(\frac{1+2i}{4}\right)}^3+7{\left(\frac{1+2i}{4}\right)}^2-\left(\frac{1+2i}{4}\right)+16$

$=\left(\frac{1+8i^3+6i(1+2i)}{64}\right)+7\left(\frac{1+4i^2+4i}{16}\right)-\left(\frac{1+2i}{4}\right)+16$

$=\left(\frac{1-8i+6i+12i^2}{64}\right)+7\left(\frac{1-4+4i}{16}\right)-\left(\frac{1+2i}{4}\right)+16$

$=\left(\frac{1-2i-12}{64}\right)+7\left(\frac{-3+4i}{16}\right)-\left(\frac{1+2i}{4}\right)+16$

$=\left(\frac{-11-2i}{64}\right)+\left(\frac{-21+28i}{16}\right)-\left(\frac{1+2i}{4}\right)+16$

$=\left(\frac{-11-2i-84+112i-16-32i+1024}{64}\right)$

$=\frac{913+78i}{64}$

Hence, the value is $\frac{913+78i}{64}$.