Given z=4(cos(x)+isin(x)), how would you express the real parts of (z-4)^2?

Question

Given z=4(cos(x)+isin(x)), how would you express the real parts of (z-4)^2?

Answer 1

Is the x real or complex? Assuming its real you substitute the expression for z straight into the second expression:

(z-4)^2

=(4(cos(x)+i sin(x))-4)^2

=4^2 ((cos(x)-1)+i sin(x))^2

=16 ((cos (x)-1)^2+2i(cos(x)-1)sin x - (sin(x))^2)

The real part of that is:

Re((z-4)^2) = 16(cos^2(x)-2cos(x)+1-(1-cos^2(x)))  

[using 1=sin^2(x)+cos^2(x) for all real x]

=16(2cos^2(x)-2cos(x))

=32(cos(x)-1)cos(x)

When x is complex you need to substitute x=u+vi into the first expression and expand the cos(x) and sin(x) using the usual trigonometric identities, and using the fact that sin(vi)=isinh(v) and cos(vi)=cosh(v), and then proceeding as above.

Answer 2

First Z - 4 = 4[ cosx + isinx -1]

               = 4[ isinx + i^2(1-cosx)]

              = 4isinT[cosT +isinT] where T = x/2.

     (z - 4)^2 = - 16 sin^2T[cosT+isinT]^2

                 -16sin^2T[cos2T - isin2T]

so the reqd real part = - 16 [cosx][sin(x/2)]^2