Question

/ Modulus of (√7-3i)^3
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Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ modulus \: of \: ( \sqrt{7} - 3i) {}^{3} \\ \\ so \: we \: know \: that \\ (a - b) {}^{3} = a {}^{3} - b {}^{3} - 3a {}^{2} b + 3ab {}^{2} \\ \\ thus \: then \: accordingly \\ \\ ( \sqrt{7} - 3i) {}^{3} = ( \sqrt{7} ) {}^{3} + (3i) {}^{3} + 3 \times \sqrt{7} \times i {}^{2} - 3 \times ( \sqrt{7} ) {}^{2} \times i \\ \\ = 7 \sqrt{7} + 27i {}^{3} + 3 \sqrt{7} \times ( - 1) - (3 \times 7i) \\ \\ = 7 \sqrt{7} + 27( - i) - 3 \sqrt{7} - 21i \\ \\ = 4 \sqrt{7} - 27i - 21i \\ \\ = 4 \sqrt{7} - 48i \\ \\ comparing \: it \: with \: a + ib \: \: and \: equating \\ real \: and \: imaginary \: parts \\ \\ we \: get \\ a = 4 \sqrt{7} \\ b = - 48$

$now \: we \: know \: that \\ \\ |z| = \sqrt{a {}^{2} + b {}^{2} } \\ \\ = \sqrt{(4 \sqrt{7} ) {}^{2} + ( - 48) {}^{2} } \\ \\ = \sqrt{112 + 2304} \\ \\ = \sqrt{2416} \\ \\ = \sqrt{16 \times 151} \\ \\ |z| = 4 \sqrt{151}$