Question

find the modulus and argument of -16/1+i√3​

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: given \: complex \: number \: is \\ \frac{ - 16}{1 + \sqrt{3}.i } \\ \\ z = \frac{ - 16}{1 + \sqrt{3} .i} \\ \\ = \frac{ - 16}{1 + \sqrt{3} .i} \: \times \: \frac{1 - \sqrt{3} i}{1 - \sqrt{3} i} \\ \\ = \frac{ - 16(1 - \sqrt{3}i) }{(1) {}^{2} - ( \sqrt{3}i) {}^{2} } \\ \\ = \frac{ - 16(1 - \sqrt{3} i)}{1 - 3i {}^{2} } \\ \\ = \frac{ - 16(1 - \sqrt{3} i)}{1 - 3( - 1)} \\ \\ = \frac{ - 16(1 - \sqrt{3}i) }{1 + 3} \\ \\ = \frac{ - 16(1 - \sqrt{3}i }{4} \\ \\ = - 4(1 - \sqrt{3} i) \\ \\ = - 4 + 4 \sqrt{3} i \\ \\ comparing \: real \: parts \: and \: imaginary \: parts \\ we \: have \: then \\ \\ a = - 4 \\ b = 4 \sqrt{ 3} \\ \\ we \: know \: that \\ \\ |z| = \sqrt{a {}^{2} + b {}^{2} } \\ \\ = \sqrt{( - 4) {}^{2} + (4 \sqrt{3} ) {}^{2} } \\ \\ = \sqrt{16 +48 } \\ \\ = \sqrt{64} \\ \\ |z| = 8$

$thus \: then \: accordingly \\ \\ arg \: (z) =θ = tan \: {}^{ - 1} \: ( \frac{b}{a} \: ) \\ \\ tan \: θ = \frac{b}{a} \\ \\ = \frac{4 \sqrt{3} }{ - 4} \\ \\ = - \sqrt{3} \\ \\ so \: we \: know \: that \\ \\ tan \: \frac{\pi}{3} = \sqrt{3} \\ \\ but \: as \: (a,b) = ( - 4 ,4 \sqrt{3} \: ) \: lies \: in \\ quadrant \: 2 \\ \\ tan \: θ = 120 =( \frac{2\pi}{3} \: ) {}^{c}$