Power of a Complex Number...2

Section 6.6

 

Apply binomial theorem:
where \(a=3\), \(b=-2i\)




substitute a and b, and expand

= \(8!/(0!(8-0)!)*3^8(-2i)^0+8!/(1!(8-1)!)3^7*(-2i)^1+8!/(2!(8-2)!)3^6(-2i)^2\)
\(+8!/(3!(8-3)!)*3^5*(-2i)^3+8!/(4!(8-4)!)*3^4*(-2i)^4+8!/(5!(8-5)!)*3^3*(-2i)^5\)
\(+8!/(6!(8-6)!)*3^2*(-2i)^6+ 8!/(7!(8-7)!)*3*(-2i)^7+ 8!/(8!(8-8)!)*3^0*(-2i)^8\)

=\(6561-34992i-81648+108864i+90720-48384i-16128+3072i+256\)

= \(-239+2856*i \)

 

There comes a time when a student must admit defeat. I give up!

Power of a Complex Number...1