2 years ago

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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Principal argument for (√3-i)^6​

Principal argument for (√3-i)^6