jj! 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
