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

1 Answer
Gió
Jun 8, 2016

I found: #z=3i=3[cos(90^@)+isin(90^@)]#

Explanation:

We can find the position of your number on the complex plane:
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We have that:
#z=3i=rho[cos(theta)+isin(theta)]# in trigonometric form, where:
#rho=# modulus, the distance from the origin to your point; in this case: #rho=3#
#theta=# angle with the horizontal Real positive semi-axis; in this case: #theta=90^@#.
So you get:
#z=3i=3[cos(90^@)+isin(90^@)]#

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