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

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Answer 1 · 2017-06-13T15:44:11

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

Consider the diagram:

enter image source here

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^@)]#