1. ## Complex analysis

Hi i need help with these 2 questions,

1) Prove that, if z=x+iy, then cotz = (sin2x-isinh2y)/(cos2y -cos2x)

i got,....cosz = cosxcoshy - isinxsinhy

and sinz = sinxcoshy + icosxsinhy

cotz = cosz/sinz

= (cosxcoshy - isinxsinhy) / sinxcoshy + icosxsinhy

Rationalizing and using double angle formula i get

cotz = (sin2x-isinh2y)/(cos2x-cosh2y)

2) By considering summation n=1 to N-1 of Z^2n and that Z=e^i@

show that summation n=1 to N-1 cos2nQ = (sinN@cos(N-1)@)/sin@,...sin@ not equal to zero

for the second one, i need an idea how to begin it....

P.S i wrote summation coz i dont know how to put that big E

2. Originally Posted by Kai
Hi i need help with these 2 questions,

1) Prove that, if z=x+iy, then cotz = (sin2x-isinh2y)/(cos2y -cos2x)

i got,....cosz = cosxcoshy - isinxsinhy

and sinz = sinxcoshy + icosxsinhy

cotz = cosz/sinz

= (cosxcoshy - isinxsinhy) / sinxcoshy + icosxsinhy

Rationalizing and using double angle formula i get

cotz = (sin2x-isinh2y)/(cos2x-cosh2y)

2) By considering summation n=1 to N-1 of Z^2n and that Z=e^i@

show that summation n=1 to N-1 cos2nQ = (sinN@cos(N-1)@)/sin@,...sin@ not equal to zero

for the second one, i need an idea how to begin it....

P.S i wrote summation coz i dont know how to put that big E
1) I think your result should be
$\cot(x+iy) = \frac{\sin 2x - i \sinh 2y}{\cosh 2y - \cos 2x}$

2) I can't follow your problem statement. I gather @ means theta, but what is Q (in cos2nQ)?

Maybe you can restate the problem?

3. its cos2n@,..anyway ive been able to solve them , thanks

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# prove of Sin z=Sinxcoshy icosxsinhy

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