Question

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

Answer 1

Take a look at the picture below and try to work out how to prove it.

Answer 2

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
\[|\cos (x+iy)|^{2} = \left( \cos (x)\cosh(y) - i \sin(x)\sinh(y) \right)^{2}\]
\[|\cos (x+iy)|^{2} = \cos^{2}(x)\cosh^{2}(y) - \sin^{2}(x) \sinh^{2}(y) - 2isin^{2}(x)\sinh^{2}(y)\]

Answer 3

and then...., what should i do ?? have you idea, guys ??
@UnkleRhaukus @sirm3d @hartnn

Answer 4

the first thing coming to my mind is to write sin^2x = (1-cos^2x)
and simplify

Answer 5

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| = \(\sqrt{a^2+b^2}\)

Answer 6

the point is, since you are squaring the magnitude, there should be no imaginary part....

Answer 7

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

Answer 8

i'm little bit confused, why \[|(\cos(x)\cosh(y) - i \sin(x)\sinh(y))|^{2} = [\cos(x)\cosh(y)]^{2} + [\sin(x)\sinh(y)]^{2}\]

Answer 9

\(|a-ib|^2=[\sqrt{a^2+b^2}]^2\)
a=cos x coshy , b= sinx sinhy

Answer 10

i think afterthat the required result is easy to get

Answer 11

ah ok.., i'll try., :),, thank you

Answer 12

ask if doubts,
and having difficulty with any other part of the question ?

Answer 13

hmm.., would u kindly help me with another questions ?? still talk about complex variable.,

Answer 14

another Q ?? taylor series one ?

Answer 15

yes..., taylor series

Answer 16

find the taylor series expansion of f (z) = z exp (2z) about z = -1, and Determine the region of convergence.

Answer 17

have idea @Roya ??

Answer 18

yes., for for trigonometry, i got it now.., thank you so much @hartnn

Answer 19

accept my apology dear @gerryliyana i was so busy . i see @hartnn solved it. tnx