FP1 2017 4 Edexcel

(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.

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