Questions on Algebra: Trigonometry answered by real tutors!

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Question 150312: Please help me write (3-2i) in polar form.
Here are the choices:
1. sqrt13(cos 33.7 degrees + i sin 33.7 degrees)
2. sqrt5(cos 326.3 degrees + i sin 326.3 degrees)
3. sqrt5(cos 33.7 degrees + i sin 33.7 degrees)
4. sqrt13(cos 326.3 degrees + i sin 326.3 degrees)
Thanks, Tami: Please help me write (3-2i) in polar form.
Here are the choices:
1. sqrt13(cos 33.7 degrees + i sin 33.7 degrees)
2. sqrt5(cos 326.3 degrees + i sin 326.3 degrees)
3. sqrt5(cos 33.7 degrees + i sin 33.7 degrees)
4. sqrt13(cos 326.3 degrees + i sin 326.3 degrees)
Thanks, Tami
Answer by Fombitz(1744) About Me  (Show Source):
You can put this solution on YOUR website!
abs(3-2i)=sqrt(3^2+2^2)=sqrt(13)
3-2i=sqrt(13)(3/sqrt(13)-2i/sqrt(13))
cos(A)=3/sqrt(13)
A=33.7 or A=360-33.7=326.3
Since, 3-2i is in the 4th quadrant, then A=326.3
3-2i=sqrt(13)(cos(326.3)+i*sin(326.3))
Looks like #4.