Question

(i) Find sq. root of complex number $z=3+4i$
(ii) Find the value of $x^3-x^2+x+46$ if $x=2+3i$

Answer 1

Part I

Square root of

$3+4i$

$(x+iy{)}^2=3+4i$

$\Rightarrow (x^2-y^2)+i\left(2xy\right)=3+4i$

fn comparing

$x^2-y^2=3$ & $xy=2$

$\Rightarrow x=2$ and $y=1$

$\therefore (2+i{)}^2=3+4i$

Part II

$x^3-x^2+x+4b=?$ if $x=2+3i$]

$x^2=(2+3i{)}^2=(2^2-3^2)+i(2\left)\right(2\left)\right(3)$

$x^2=-5+12i$

$x^3=(-5+12i)(2+3i)$

$=(-10-15i+24i-3b)$

$x^3=[-46+9i]$

$x^3-x^2+x+46=$

$\Rightarrow [-46+9i]-[-5+12i]+[2+3i]+4b$

$\Rightarrow -46+9i+5-12i+2+3i+4b$

$\Rightarrow 7+12i-12i$

$\Rightarrow 7$

$\therefore x^3-x^2+x+4b=7$ if $x=2+3i$