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How do you write #2 - 3i# in trigonometric form?

1 Answer

Jan 27, 2016

#2-3i = sqrt(13)(cos theta +isin theta)#

Explanation:

See Trigonometric form for a demonstration of how to do this.

In order to write a complex number #2 - 3i# in trigonometric form we need to find the modulus #z#, where #z = a +bi = |z|(cos theta + i sin theta)#

#z=sqrt(a^2+b^2) = sqrt(2^2 +(-3)^2) =sqrt(13)#

Then #2-3i = sqrt(13)(cos theta +isin theta)#

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