## gerryliyana 2 years ago hi guys..., i have a complex trigonometry question.

1. gerryliyana

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

2. gerryliyana

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

3. gerryliyana

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

4. hartnn

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

5. hartnn

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| = $$\sqrt{a^2+b^2}$$

6. hartnn

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

7. hartnn

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

8. gerryliyana

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

9. hartnn

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

10. hartnn

i think afterthat the required result is easy to get

11. gerryliyana

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

12. hartnn

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

13. gerryliyana

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

14. hartnn

another Q ?? taylor series one ?

15. gerryliyana

yes..., taylor series

16. gerryliyana

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

17. gerryliyana

have idea @Roya ??

18. gerryliyana

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

19. Roya

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