Quod Erat Demonstrandum

2008/04/21

Complex trigonometric functions

Filed under: University Mathematics — johnmayhk @ 1:03 下午
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F.5 student Hoover found it interesting to know some properties of complex trigonometric functions. He asked if the following is still true?

\sin^2(z) + \cos^2(z) = 1

for any complex number z.

The answer is affirmative. Here is a verification.

Start with the famous Euler’s formula,

e^{iz} = \cos(z) + i\sin(z)

Replace z by -z, yields

e^{-iz} = \cos(z) - i\sin(z)

By elimination from the equations above, we have

\cos(z) = \frac{e^{iz} + e^{-iz}}{2}
\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}

Now, it is a piece of cake to obtain

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

Could you try to verify another usual properties of real trigonometric functions to see if they are still applicable to the cases in complex numbers? Like say

\cos(2z) = \cos^2(z) - \sin^2(z)

*\sin(-z) = -\sin(z) may be verified by the power series of the sine function.

5 則迴響 »

  1. 睇過d書講imaginary number既出現係因為處理cubic equation

    所以我有個問題想問下John Sir,complex number在實際應用的層面有幾大?

    我只知道在solve differential equation等的applied maths會用到,另外亦都聽過d係讀engine,當處理某些問題也會用到complex number

    迴響 由 Justin — 2008/04/21 @ 1:42 下午 | 回覆

  2. 讀 civil engineering 時,其中一科是 electronics。教『三相供電』時,就是把電壓或電流以 complex numbers 表示。相信是為了方便計算而已(即不用 complex numbers 也可)。確實,在工程科目中,complex numbers 是非常有用的。看看這個簡介:

    http://en.wikipedia.org/wiki/Complex_number#Applications

    迴響 由 johnmayhk — 2008/04/21 @ 3:13 下午 | 回覆

  3. Here is another nice fact, suppose f, and g are analytic functions in D (here D is a simply connected domain, for example, we may take D to be the open unit disc or the entire complex plane)

    Suppose f^2 + g^2 = 1 for z in D.

    Then we have f = cos(ih(z)), g = sin(-ih(z)) for some analytic function h in D.

    Here is a short proof, we have (f + ig)(f – ig) = 1, and hence f + ig never vanishes in D, thus f + ig = e^{h(z)} for some analytic function h in D. Likewise, f – ig = 1/(f + ig) = e^{-h(z)}.
    Therefore, f = [e^{h(z)} + e^{-h(z)}]/2 = cos(ih(z)), and g = sin(-ih(z)). Done.

    Hence, sin and cos are essentially the only analytic functions that satisfy f^2 + g^2 = 1.

    迴響 由 koopa — 2008/04/22 @ 1:20 下午 | 回覆

  4. Hi, this is the first time I visit your site, and it’s really nice! Just to share another idea on how to see that sin^2(z) + cos^2(z) = 1 holds even for complex numbers: one just needs to observe that both sides of the equation are analytic functions on the entire complex plane, and that the identity holds for real numbers z. Then apply the identity theorem for analytic functions. Almost all common trigonometric identities extend to complex arguments this way.

    迴響 由 Polam — 2008/04/25 @ 6:54 上午 | 回覆

  5. Thank you very much for the nice facts given by Koopa and Polam! Yes, the identity theorem should be taught at the very beginning of the complex analysis. I just overlook that. Thank you for reminding me that complex analysis is fun.

    迴響 由 johnmayhk — 2008/04/25 @ 3:32 下午 | 回覆


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