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Math Help - Complex numbers.

  1. #1
    Super Member Showcase_22's Avatar
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    Complex numbers.

    The question is:

    Let z=\frac{\pi}{6}+iln2. Write e^{iz} in the form a+bi.

    (careful! This is a trick question.)


    What I did:

    e^{iz}=e^{\frac{i \pi}{6}-ln2}

    e^{iz}=e^{\frac{i\pi}{6}}e^{-ln2}

    e^{iz}=\frac {1}{2} e^{\frac{i\pi}{6}}

    e^{iz}=(\frac{1}{2})(cos(\frac{\pi}{6}) + i sin(\frac{\pi}{6}))

    e^{iz}=(\frac{1}{2})(\frac{\sqrt 3}{2}+ i \frac{1}{2})

    e^{iz}=(\frac{1}{4})(\sqrt3+ i)

    I think that's right. It's just that when a lecturer has taken time out to write "this is a trick question" next to it, it implies it is much harder than it looks.

    So is this the right answer?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    The question is:

    Let z=\frac{\pi}{6}+iln2. Write e^{iz} in the form a+bi.

    (careful! This is a trick question.)

    What I did:

    e^{iz}=e^{\frac{i \pi}{6}-ln2}

    e^{iz}=e^{\frac{i\pi}{6}}e^{-ln2}

    e^{iz}=\frac {1}{2} e^{\frac{i\pi}{6}}

    e^{iz}=(\frac{1}{2})(cos(\frac{\pi}{6}) + i sin(\frac{\pi}{6}))

    e^{iz}=(\frac{1}{2})(\frac{\sqrt 3}{2}+ i \frac{1}{2})

    e^{iz}=(\frac{1}{4})(\sqrt3+ i)

    I think that's right. It's just that when a lecturer has taken time out to write "this is a trick question" next to it, it implies it is much harder than it looks.

    So is this the right answer?
    Well let's see...

    If z = x + iy then e^z = e^x \ {\rm{cis}} (y).
    Last edited by CaptainBlack; October 2nd 2008 at 10:45 PM. Reason: correct LaTeX
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  3. #3
    Super Member Showcase_22's Avatar
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    ahhhh! The last bit you wrote ended up with a Latex error!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Showcase_22 View Post
    ahhhh! The last bit you wrote ended up with a Latex error!
    Corrected, but what it says you knew already!

    RonL
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    Grand Panjandrum
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    Quote Originally Posted by Showcase_22 View Post
    The question is:

    Let z=\frac{\pi}{6}+iln2. Write e^{iz} in the form a+bi.

    (careful! This is a trick question.)

    What I did:

    e^{iz}=e^{\frac{i \pi}{6}-ln2}

    e^{iz}=e^{\frac{i\pi}{6}}e^{-ln2}

    e^{iz}=\frac {1}{2} e^{\frac{i\pi}{6}}

    e^{iz}=(\frac{1}{2})(cos(\frac{\pi}{6}) + i sin(\frac{\pi}{6}))

    e^{iz}=(\frac{1}{2})(\frac{\sqrt 3}{2}+ i \frac{1}{2})

    e^{iz}=(\frac{1}{4})(\sqrt3+ i)

    I think that's right. It's just that when a lecturer has taken time out to write "this is a trick question" next to it, it implies it is much harder than it looks.

    So is this the right answer?
    Well it looks OK, but we can just do a numerical check:

    Code:
    >I
                           0+1i 
    >z=pi/6+I*log(2)
             0.523599+0.693147i 
    >exp(I*z)
                 0.433013+0.25i 
    >
    >
    >(0.25)*(sqrt(3)+I)
                 0.433013+0.25i 
    >
    RonL
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  6. #6
    Super Member Showcase_22's Avatar
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    Corrected, but what it says you knew already!
    Actually, i'm not sure I do :s

    What does e^z=e^x cis (y) mean? More specifically, what does "cis" stand for?

    Well it looks OK, but we can just do a numerical check:

    >exp(I*z)
    0.433013+0.25i
    How did you work out that bit? Can your calculator work out complex numbers?
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  7. #7
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    Quote Originally Posted by Showcase_22 View Post
    Actually, i'm not sure I do :s

    What does e^z=e^x cis (y) mean? More specifically, what does "cis" stand for?
    cis(y) is a shorthand way of writing \cos{y} + i\sin{y}, which is a complex number of magnitude 1 written in polar co-ordinates.

    By definition of z = x + iy, we have

    e^z = e^{x+iy} = e^x e^{iy} = e^x cis(y)

    because, by Euler's formula

    e^{i\theta} = cis(\theta)
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Showcase_22 View Post
    Actually, i'm not sure I do :s

    What does e^z=e^x cis (y) mean? More specifically, what does "cis" stand for?
    It actualy means very little, it is a relativly recently invented notation for:

    \cos(x)+{\bold{i}}\sin(x)

    but such a notation is unneccessary, and redundant.


    How did you work out that bit? Can your calculator work out complex numbers?
    Well mine can, yes

    RonL
    Last edited by CaptainBlack; October 4th 2008 at 01:57 AM.
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  9. #9
    Super Member Showcase_22's Avatar
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    oh right!

    That cis notation looks redundant but it is shorter than writing cos x+i sinx. I might start using it on my whiteboard since I always run out of space on it!

    My calculator can't do complex numbers. Well I don't think it can!

    Does a TI-84 have a complex number setting? I haven't really used it and I didn't take the manual with me to university.

    Also, for a trick question, that wasn't that bad. I guess lecturers write that there to freak students out!
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