#j#! I'm a EE again!
I know we're asked about (iv) but let's do the whole thing. I won't look what appears to be the answer until the end.
#f(z) = 2z^4 - 9z^3 + Az^2 + Bz - 26, quad A, B# real
We're given one root #gamma =3+2j# and the real coefficients mean the complex roots come in conjugate pairs, so another root is #delta = bar{gamma} = 3-2j.# #gamma + delta = 6# and the product of conjugates is the squared magnitude, #gamma delta=3^2+2^2=13.#
So we have
#(z-gamma)(z-delta) =z^2 - (gamma+delta) z + gamma delta = z^2 -6z + 13#
We'll come back to that.
Let's derive and apply the relevant Viete's formulas for our case. #1/2 f(z)# is a monic polynomial with the same zeros as #f#. We sorta know its factorization because we've named the roots:
#1/2 f(z) = z^4 + 9/2 z^3 + A/2 z^2 + B/2 z - 13 = (z- alpha)(z-beta)(z-gamma)(z-delta) #
We see the constant term
#-13 = (-alpha)(-beta)(-gamma)(-delta) = alpha beta gamma delta = 13 alpha beta# so
#alpha beta = -1#
The cubic term we see is
#-9/2 z^3 = (-alpha - beta - gamma - delta) z^3#
#9/2 = alpha + beta + gamma + delta = alpha + beta + 6#
#alpha + beta = -3/2#
So now we have
#0 = (z-alpha)(z-beta)=z^2-(alpha+beta)z + alpha beta = z^2+3/2 z - 1 #
# 0 = 2z^2 + 3 z - 2 #
#0= (2z - 1)(z + 2)#
#z = 1/2 or z=-2#
#alpha>beta# so #alpha = 1/2, beta=-2#
Our original #f(z)# must be
#f(z) = 2(z- alpha)(z-beta)(z-gamma)(z-delta) #
#f(z)=(2z^2 + 3 z - 2)(z^2 -6z + 13)#
#f(z) = 2 z^4 - 9 z^3 + 6 z^2 + 51 z - 26#
#A=6#
#B=51#
OK, finally to part iv.
If we know #f(z)=0# then
#f(j/j z) = 0#
#f((jz)/j) = 0#
So if #w=jz#
#f(w/j) = 0#
So we get
#w = jalpha, jbeta, jgamma, jdelta#
#w = j/2, -2j, -2+3j, 2+3j#
