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

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

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

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

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

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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.3so 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

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

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

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

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

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

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

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

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

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

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

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

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