How do you write the complex number in trigonometric form 8+3i?

1 Answer
Jun 13, 2017

I got: z=r[cos(theta)+isin(theta)]=8.54[cos(20.56^@)+isin(20.56^@)]

Explanation:

Consider the diagram:

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From Pythagoras:

r=sqrt(8^2+3^2)=8.54

From Trigonometry:

theta=arctan(3/8)=20.56^@

So:

z=8+3i=r[cos(theta)+isin(theta)]=8.54[cos(20.56^@)+isin(20.56^@)]