Respuesta :

naomi8180

Answer:

z = (1 + i)6 =³√2(cos π4+ isinπ4)´6= 8 µcos3π2+ isin3π2)¶= −8i. Hence |z| = 8 and ¯z = 8i. z = (1 + i)

I think I am not real sure

LammettHash

In polar form, we have

[tex]1+i = \sqrt2 e^{i\pi/4}[/tex]

Then by DeMoivre's theorem,

[tex](1+i)^6 = \left(\sqrt2\right)^6 e^{i\,6\pi/4} = 2^3 e^{i\,3\pi/2} = -8i[/tex]

Then

[tex]|z| = |-8i| = \boxed{8}[/tex]

and

[tex]\bar z = \overline{-8i} = \boxed{8i}[/tex]

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