Question

Find the square roots of a complex number (-7-24i)

Answer 1

Convert complex number to polar no,s

−7−24i=25cos(−π+atan(247))+25isin(−π+atan(247))

According to the De Moivre's Formula, all n-th roots of a complex number r(cos(θ)+isin(θ)) are given by r√n(cos(θ+2πkn)+isin(θ+2πkn)), k=0..n−1¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

We have that r=25, θ=−π+atan(247), n=2.

Thus,

k=0: 25−−√2(cos((−π+atan(247))+2π⋅02)+isin((−π+atan(247))+2π⋅02))=5(cos(−π2+atan(247)2)+isin(−π2+atan(247)2))=5sin(atan(247)2)−5icos(atan(247)2)

k=1: 25−−√2(cos((−π+atan(247))+2π⋅12)+isin((−π+atan(247))+2π⋅12))=5(cos(atan(247)2+π2)+isin(atan(247)2+π2))=−5sin(atan(247)2)+5icos(atan(247)2)

ANSWER

−7−24i−−−−−−−√2=5sin(atan(247)2)−5icos(atan(247)2)≈3.0−4.0i

−7−24i−−−−−−−√2=−5sin(atan(247)2)+5icos(atan(247)2)≈−3.0+4.0i

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Answer 2

ANSWER:-

$let \sqrt{ - 7 - 24i} = a + ib$

$- 7 - 24i = (a + ib {)}^{2} = {a}^{2} - {b}^{2} + 2iab$

$comparing \: coeffiecient \: we \: get$

${a}^{2} - {b}^{2} = - 7 \: \: and \: \: 2ab = - 24$

$ab = - 12$

$b = \frac{ - 12}{a}$

${a}^{2} - \frac{144}{ {a}^{2} } = - 7$

${a}^{2} + 7 {a}^{2} - 144 = 0$

$= > ( {a}^{2} - 9)( {a}^{2} + 16) = 0$

$Hence, {a}^{2} + 16≠0 \: \: \: so, {a}^{2} = 9$

$a = ±3$

$a = \frac{ - 12}{a} = ±4$

$for \: a = 3,b = - 4$

$a = - 3,b = - 4$

$so, = \sqrt{ - 7 - 24i} = ±(3 - 4i)$

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