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hi guys..., i have a complex trigonometry question.

Trigonometry
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Take a look at the picture below and try to work out how to prove it.
1 Attachment
i've tried ..., using all identity above.., i have \[|\cos z|^{2} = |\cos (x+iy)|^{2} = \cos (x) \cos (iy) - \sin (x) \sin (iy)\] because of: \(\cos (iy) = \cosh(y) \) and \(\sin (iy) = i \sinh (y)\), then i have \[|\cos (x+iy)|^{2} = \left( \cos (x)\cosh(y) - i \sin(x)\sinh(y) \right)^{2}\] \[|\cos (x+iy)|^{2} = \cos^{2}(x)\cosh^{2}(y) - \sin^{2}(x) \sinh^{2}(y) - 2isin^{2}(x)\sinh^{2}(y)\]
and then...., what should i do ?? have you idea, guys ?? @UnkleRhaukus @sirm3d @hartnn

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Other answers:

the first thing coming to my mind is to write sin^2x = (1-cos^2x) and simplify
also, i think you missed the |...| shouldn't it be \(|\cos z|^{2} = |\cos (x+iy)|^{2} = |\cos (x) \cos (iy) - \sin (x) \sin (iy)|^2 \\= | \left( \cos (x)\cosh(y) - i \sin(x)\sinh(y) \right)|^{2}= [\cos (x)\cosh(y)]^2 +[ \sin(x)\sinh(y)]^2\) because |a+ib| = \(\sqrt{a^2+b^2}\)
the point is, since you are squaring the magnitude, there should be no imaginary part....
so you should've got, \(|\cos z|^{2} = [\cos (x)\cosh(y)]^2 +[ \sin(x)\sinh(y)]^2\) did you understand how ? now write sin^2x = 1- cos^2 x
i'm little bit confused, why \[|(\cos(x)\cosh(y) - i \sin(x)\sinh(y))|^{2} = [\cos(x)\cosh(y)]^{2} + [\sin(x)\sinh(y)]^{2}\]
\(|a-ib|^2=[\sqrt{a^2+b^2}]^2\) a=cos x coshy , b= sinx sinhy
i think afterthat the required result is easy to get
ah ok.., i'll try., :),, thank you
ask if doubts, and having difficulty with any other part of the question ?
hmm.., would u kindly help me with another questions ?? still talk about complex variable.,
another Q ?? taylor series one ?
yes..., taylor series
find the taylor series expansion of f (z) = z exp (2z) about z = -1, and Determine the region of convergence.
have idea @Roya ??
yes., for for trigonometry, i got it now.., thank you so much @hartnn
accept my apology dear @gerryliyana i was so busy . i see @hartnn solved it. tnx

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