Question

(2i)^3 - (2i)^2 + 2i -1 CALCULATE AND ANSWER IN CARTESIAN form.

Answer 1

\[8\iota^2\iota-4\iota^2+2\iota-1=-8\iota+4+2\iota-1=3-6\iota \]
\[suppose ~x=rcos \theta=3,y=r \sin \theta=-6,\square~and~add\]
\[x^2+y^2=r^2\left( \cos ^2\theta+\sin ^2 \theta \right)=3^2+\left( -6 \right)^2=9+36=45\]

Answer 2

or it can be like this
\[x+\iota y=3-6 \iota \iota,\left| x+\iota y \right|=\left| 3-6\iota \right|\]
\[\sqrt{x^2+y^2}=\sqrt{3^2+\left( -6 \right)^2},~or~x^2+y^2=9+36=45\]