Question

if w =(7 -z)/ (1- z^2) where z = 1+2i, then find [w].

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ value \: of \: ω \\ \\ given : \\ \\ ω = \frac{7 - z}{1 - z {}^{2} } \\ \\ z = 1 + 2i \\ \\ so \: then \\ \\ ω = \frac{7 - z}{1 - z {}^{2} } \\ \\ = \frac{7 - (1 + 2i)}{1 - (1 + 2i) {}^{2} } \\ \\ = \frac{7 - 1 - 2i}{1 - (1 + 4i {}^{2} + 4i) } \\ \\ = \frac{6 - 2i}{1 - 1 - 4i {}^{2} - 4i } \\ \\ = \frac{6 - 2i}{ - 4( - 1) - 4i} \\ \\ = \frac{2(3 - i)}{4 - 4i} \\ \\ = \frac{2(3 - i)}{4(1 - i)} \\ \\ = \frac{3 - i}{2(1 - i)}$

$= \frac{3 - i}{2(1 - i)} \times \frac{1 + i}{1 + i} \\ \\ = \frac{(3 - i)(1 + i)}{2(1 - i)(1 + i)} \\ \\ = \frac{3 + 3i - i - i {}^{2} }{ 2(1 - i {}^{2} ) } \\ \\ = \frac{3 - ( - 1) + 2i}{2(1 - ( - 1))} \\ \\ = \frac{3 + 1 + 2i}{2(1 + 1)} \\ \\ = \frac{4 + 2i}{2 \times 2} \\ \\ = \frac{2(2 + i)}{4} \\ \\ = \frac{2 + i}{2} \\ \\ ω = a + ib = 1 + \frac{1}{2} i$