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gerryliyanaBest ResponseYou've already chosen the best response.0
Take a look at the picture below and try to work out how to prove it.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
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 \[\cos (x+iy)^{2} = \left( \cos (x)\cosh(y)  i \sin(x)\sinh(y) \right)^{2}\] \[\cos (x+iy)^{2} = \cos^{2}(x)\cosh^{2}(y)  \sin^{2}(x) \sinh^{2}(y)  2isin^{2}(x)\sinh^{2}(y)\]
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
and then...., what should i do ?? have you idea, guys ?? @UnkleRhaukus @sirm3d @hartnn
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
the first thing coming to my mind is to write sin^2x = (1cos^2x) and simplify
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
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 = \(\sqrt{a^2+b^2}\)
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
the point is, since you are squaring the magnitude, there should be no imaginary part....
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
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
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
i'm little bit confused, why \[(\cos(x)\cosh(y)  i \sin(x)\sinh(y))^{2} = [\cos(x)\cosh(y)]^{2} + [\sin(x)\sinh(y)]^{2}\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
\(aib^2=[\sqrt{a^2+b^2}]^2\) a=cos x coshy , b= sinx sinhy
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
i think afterthat the required result is easy to get
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
ah ok.., i'll try., :),, thank you
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
ask if doubts, and having difficulty with any other part of the question ?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
hmm.., would u kindly help me with another questions ?? still talk about complex variable.,
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
another Q ?? taylor series one ?
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
yes..., taylor series
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
find the taylor series expansion of f (z) = z exp (2z) about z = 1, and Determine the region of convergence.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
have idea @Roya ??
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
yes., for for trigonometry, i got it now.., thank you so much @hartnn
 one year ago

RoyaBest ResponseYou've already chosen the best response.0
accept my apology dear @gerryliyana i was so busy . i see @hartnn solved it. tnx
 one year ago
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