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Trig chart for the complex plane

  1. Jul 28, 2005 #1
    i was trying to make a trig chart for the complex plane and accindently found 0>1. at pi/4, the hypotenuse is 0, but the real leg is 1. how is that? imaginary trig must be fun.

    at [tex]\theta=\frac{\pi}{4}[/tex], i got:

    sin=undefined
    cos=undefined
    tan=i
    csc=0
    sec=0
    cot=-i

    it is in the first quadrant and cot is negative...something is wonky
     
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  3. Jul 28, 2005 #2

    James R

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    What's complex about trig functions of [itex]\theta = \pi/4[/itex]? It is not clear to me what you're doing. Hypotenuse of what? Real leg of what?
     
  4. Jul 28, 2005 #3
    the legs are of the triangle. i thought that was implied, sorry.
     
  5. Jul 28, 2005 #4

    James R

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    So, are you talking about points on the unit circle in the complex plane?

    By definition, all points on that circle have modulus 1. The real part of any point on the circle is given by [itex]\cos \theta[/itex], where [itex]\theta[/itex] is the angle between the real axis and the line joining the origin to the point on the circle.

    [itex]\cos \pi/4 = 1/\sqrt{2}[/itex]

    Is this what you're talking about, or something else?
     
  6. Jul 28, 2005 #5
    i drew the triangle and got it to be undefined. [tex]\frac{1}{\sqrt{2}}[/tex] is what i get on a real plane, but not on a complex one.
     
  7. Jul 28, 2005 #6

    James R

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    Please explain exactly what you are doing in your calculation.
     
  8. Jul 28, 2005 #7
    i drew the triangle like i would on a real plane and found the trig funtions for pi/4
     
  9. Jul 28, 2005 #8

    James R

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    So, what did you do differently on the complex plane compared to the real (x,y) plane?

    The value of [itex]\cos \pi/4[/itex] never changes.

    Also, you talk about having a hypotenuse of zero with a real leg of 1. Are you talking about a right-angled triangle? Because any triangle with a non-zero side adjacent to the right angle will have a non-zero hypotenuse - otherwise it isn't a right angled triangle.

    Can you post your diagram?
     
    Last edited: Jul 28, 2005
  10. Jul 28, 2005 #9
    i don't have a scanner. i didn't do anything differently. i just used a complex plane instead of a real one.
     
  11. Jul 28, 2005 #10

    James R

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    If you didn't do anything differently, then there's no way you could get a different answer.

    I guess we're done here, then.
     
  12. Jul 28, 2005 #11
    yes there is...complex plane is not the real plane...
     
  13. Jul 28, 2005 #12

    James R

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    I'm sorry, but you don't seem to be able to explain yourself clearly enough for me to understand you. Maybe somebody else will come along who will know what you're talking about.

    Good luck.
     
  14. Jul 28, 2005 #13
    i don't understand what you don't understand.

    i drew a triangle in the first quadrant with a central angle being pi/4. when i found the lenth of the hypotenuse, it was 0. what is there not to understand?
     
    Last edited: Jul 28, 2005
  15. Jul 28, 2005 #14
    How did you obtain the value "0" then? Clearly the hypotenuse as shown in your drawing was of non-zero length, so you did some sort of calculation to obtain that value.
     
  16. Jul 28, 2005 #15
    no your answer isn't correct. unless your understanding of complex or perhap hypoteneuse is muttle you can't get an answer of 0. James already stated the modulus is 1...the lenght of boths sides of your right angle are invsqrt(2)-> sqr then add you get 1...so your lenght is 1 in complex OR real...

    How on earth did you get the hyp =0?
    REAL: invsqrt(2)*i +invsqrt(2)*j
    CMPLX: invsqrt(2)*I +invsqrt(2)*R
    the coordinates are identical, thus the lengths of the sides are the same length.

    lenght of a complex vector is its modulus. sqrt(a^2+b^2)
     
  17. Jul 28, 2005 #16

    Integral

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    Since the complex plane is isomorphic to R2, I share James confusion about what you are doing. How did you come up with a zero length hypotenuse? IIRC, the hypotenuse is always 1 when working in a unit circle. It does not matter if you do it in the complex plane or R2
     
  18. Jul 28, 2005 #17

    arildno

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    yourdaddyspoggystick:

    Remember that the squared modulus of a complex number z is given by zz', where z' is the complex conjugate of z. It is not given by zz; perhaps it was here that you erred?
     
  19. Jul 28, 2005 #18

    matt grime

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    So, what he appears to be doing, worknig in the first uadrant:

    take the points (0,0) (x,0) (x,yi) where x^2+y^2=1 and take the triangle these define.

    he now chooses to define the trig functions as the ratios such as yi/x etc and believes that the pythagorus allows us to deduce that x^2+(yi)^2 is the length of the hypoteneuse. since those are not the definitino of complex trig funcions, or the "length of a hypoteneuse" we can see the mistake.
     
  20. Jul 28, 2005 #19

    arildno

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    Matt Grime's suggestion seems to be the most likely explanation of the pogostick's mistake.
     
  21. Jul 28, 2005 #20
    Given a right angled triangle with sides a and b, the angle is given by arctan(b/a), and then the sine and cosine can be calculated. This still makes sense if a and b are complex numbers. However there seems to be a problem with a=1 b=i, since I think arctan(z) in singular at z=i.

    Of course for a complex number z, the angle arg(z) is totally different from the 'angle' arctan(z) (which will be a complex number). Also arg(i) =[tex] \pi/2[/tex] not [tex] \pi/4[/tex]
     
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