Question

If $x=1+2i$, then find the value of $x^3+x^2-x+22wherei=\sqrt{-1}$.

Answer 1

Given ,

$x=1+2i$

$Putx=1+2i,in{x}^3+x^2-x+22$

We get

$x^3+x^2-x+22={(1+2i)}^3+{(1+2i)}^2-(1+2i)+22$

$=1^3+8i^3+{3.1}^{2}2i+\mathrm{3.1.}{\left(2i\right)}^2+(1^2+4i^2+4i)+22$

$=23+8i^3+16i^2+10i$

Given ,

$i=\sqrt{-i}$

$=23+8i.i^2+16i^2+10i$

$=23+8i(-1)+16(-1)+10i$

$=23-8i-16+10i$

$=7+2i$