Question

simplify-2i(3+i)(2+4i)(1+i) and obtain the modulus of that complex number​

Answer 1

Answer:

hey here is your answer

pls mark it as brainliest

Step-by-step explanation:

$so \: here \: \\ - 2i(3 + i)(2 + 4i)(1 + i) \\ \\ = - 2i(3 + i) \times 2(1 + 2i)(1 + i) \\ \\ = - 4i(3 + i)(1 + 2i)(1 + i) \\ \\ = - 4i(3 + i)(1 + i + 2i + 2i {}^{2} ) \\ \\ = ( - 12i - 4i {}^{2} )(1 + 3i + 2i {}^{2} ) \\ \\ = ( - 12i - 4( - 1)) \times (1 + 3i + 2( - 1)) \\ \\ = ( - 12i + 4)(3i - 1) \\ \\ = - 36i {}^{2} + 12i + 12i - 4 \\ \\ = - 36( - 1) + 24i - 4 \\ \\ = 36 - 4 + 24i \\ \\ = 32 + 24i \\ \\ so \: comparing \: this \: complex \: number \: with \: a + ib \\ we \: get \\ \\ a = 32 \\ b = 24 \\ \\ so \: hence \: then \\ |z| = \sqrt{a {}^{2} + b {}^{2} } \\ \\ = \sqrt{(32) {}^{2} + (24) {}^{2} } \\ \\ = \sqrt{1024 + 576} \\ \\ = \sqrt{1600} \\ \\ = 40 \\ \\ |z| = 40$