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Finding polar form of complex number

  • Thread starter astrololo
  • Start date
  • #1
200
3

Homework Statement


I have the following complex numbers : -3,18 +4,19i
I must put it in polar form.

Homework Equations


r=(a^2+b^2)^(1/2)
cos x = a/r
sin x = b/r

The Attempt at a Solution



I was able to find with cos x = a/r that the x = 127,20

But when I do it with sin x = b/r I obtain like 52 degrees. I know that I Must obtain 127,20 for BOTH. Why isnt it working ?
 

Answers and Replies

  • #2
FactChecker
Science Advisor
Gold Member
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The sin of both those angles are the same, so you must decide which is correct. That is the angle that has the correct a,b values. 52 degrees would be at (3.18, 4.19) and 127.2 degrees is at (-3.18, 4.19).
 
  • #3
200
3
The sin of both those angles are the same, so you must decide which is correct. That is the angle that has the correct a,b values. 52 degrees would be at (3.18, 4.19) and 127.2 degrees is at (-3.18, 4.19).
Oh ok so its normal that I obtain two different values. Ok then, thank you!
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Yes, There are always two values of [itex]\theta[/itex] in the interval 0 to [itex]2\pi[/itex] that have the same [itex]sin(\theta)[/itex]. But you still have to determine which is correct for the specific problem- the two different values, [itex]\theta[/itex] have different values for [itex]cos(\theta)[/itex].
 

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