(ia)
We have:
\begin{align*}w &= \frac {(p - 4i) (2 + 3i)} {(2 - 3i) (2 + 3i)} \\ & = \frac {2p - 8i + 3ip + 12} {2^2 + 3^2} \\ & = \frac {2p + 12 + i(3p - 8)} {13} \\ & = \left(\frac {2 p + 12} {13}\right) + i \left(\frac {3p - 8} {13}\right)\end{align*}
(ib)
If \arg w = \dfrac \pi 4, then the real and imaginary parts are equal and positive, so:
3p - 8 = 2p + 12
So p = 20.
(ii)
Note that:
\begin{align*}|z|^2 &= 45^2 \\ & = |(1 - \lambda i)|^2|(4 + 3i)|^2 \\ & = 25(1 + \lambda^2)\end{align*}
So:
1 + \lambda^2 = 81
Then:
\lambda = \pm \sqrt {80} = \pm 4 \sqrt 5.
1 Like
