Question

1. resolve: |(1+i)(2-i)(1-i)|^2
gente me ajudem eh análise complexa.

2. determine a parte real do nr imaginario: Z=(1+i)^12.

3. se Z= raiz de 3-i determine Z^9

Answer 1

1.

$|(1+i)(2-i)(1-i)|^{2}\\|(1+i)(1-i)(2-i)|^{2}\\|(1^{2}-i^{2})(2-i)|^{2}\\|(1-[-1])(2-i)|^{2}\\|(1+1)(2-i)|^{2}\\|2(2-i)|^{2}\\|4-2i|^{2}$

O módulo de um número complexo é dado por:

$|z|=\sqrt{(parte~real)^{2}+(parte~imaginaria)^{2}}\\|4-2i|=\sqrt{4^{2}+(-2)^{2}}\\|4-2i|=\sqrt{16+4}\\|4-2i|=\sqrt{20}\\|4-2i|^{2}=\sqrt{20}^{2}\\|4-2i|^{2}=20$

$\boxed{\boxed{|(1+i)(2-i)(1-i)|^{2}=20}}$
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2.

$z=(1+i)^{12}\\z=[(1+i)^{2}]^{6}$

$(1+i)^{2}=1^{2}+2*1*i+i^{2}\\(1+i)^{2}=1+2i-1\\(1+i)^{2}=2i$

$z=[(1+i)^{2}]^{6}\\z=(2i)^{2}\\z=2^{6}i^{6}\\z=64(i^{2})^{3}\\z=64(-1)^{3}\\z=64(-1)\\z=-64$

Parte real: - 64
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3.

$z=\sqrt{3}-i\\z^{9}=(\sqrt{3}-i)^{9}\\z^{9}=[(\sqrt{3}-i)^{3}]^{3}$

$(\sqrt{3}-i)^{3}=(\sqrt{3})^{3}-3.\sqrt{3}^{2}.i+3.\sqrt{3}.i^{2}-i^{3}\\(\sqrt{3}-i)^{3}=\sqrt{3}^{2}\sqrt{3}-3.3.i+3\sqrt{3}.(-1)-(-i)\\(\sqrt{3}-i)^{3}=3\sqrt{3}-9i-3\sqrt{3}+i\\(\sqrt{3}-i)^{3}=-9i+i\\(\sqrt{3}-i)^{3}=-8i$

$z^{9}=[(\sqrt{3}-1)^{3}]^{3}\\z^{9}=(-8i)^{3}\\z^{9}=(-8)^{3}i^{3}\\z^{9}=-512(-i)\\z^{9}=512i$