How do you write the complex number #z = 6(cos 45° + i sin 45°) # in standard form?

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How do you write the complex number #z = 6(cos 45° + i sin 45°) # in standard form?

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Answer 1 · 2016-02-21T13:37:05

I found: #3sqrt(2)+3sqrt(2)i#

Explanation:

We need to evaluate the values of #sin# and #cos# and multiply by #6#:
#z=6(cos(45^@)+isin(45^@))=6(sqrt(2)/2+isqrt(2)/2)=#
#=cancel(6)^3(sqrt(2)/cancel(2)+isqrt(2)/cancel(2))=3sqrt(2)+3sqrt(2)i#

which is in the form: #a+bi#

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How do you write the complex number #z = 6(cos 45° + i sin 45°) # in standard form?

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Explanation:

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