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galina1969 [7]
2 years ago
7

Principal argument for (√3-i)^6​

Mathematics
1 answer:
True [87]2 years ago
4 0

Answer:

r= \sqrt{(\sqrt{3})^{6} +(1)^{2}  } =\sqrt{4} =2

(2)^5(cos(330.6)+isin(330.6))

=64(Cos1350+iSin1350)

64(cos180+isin180)

=64(-1+0)= -64  

Step-by-step explanation:

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