Problem regarding complex numbers

[sankalpmittal]

Homework Statement

If m and x are two real numbers where m ε Integers, then e^2micot-1x{(xi+1)/(xi-1)}^m, (where i=√(-1)) is equal to :

(a) cos(x) + isin(x)
(b) m/2
(c) 1
(d) (m+1)/2

Homework Equations

The Attempt at a Solution

I seriously have no clear cut idea of how to proceed. I used this technique,

Since m is an integer, then I put m=0, and got the correct answer. :p

But I want a procedure, not a hit and trial method, of how to proceed.

I can write above as

{cos (2mcot^-1x) + isin(2mcot^-1x)}{(xi+1)/(xi-1)}^m

But how to proceed after this ? Hints will do..

Please help !

Thanks in advance...

 

[Ray Vickson]

Homework Statement

If m and x are two real numbers where m ε Integers, then e^2micot-1x{(xi+1)/(xi-1)}^m, (where i=√(-1)) is equal to :

(a) cos(x) + isin(x)
(b) m/2
(c) 1
(d) (m+1)/2

Homework Equations

The Attempt at a Solution

I seriously have no clear cut idea of how to proceed. I used this technique,

Since m is an integer, then I put m=0, and got the correct answer. :p

But I want a procedure, not a hit and trial method, of how to proceed.

I can write above as

{cos (2mcot^-1x) + isin(2mcot^-1x)}{(xi+1)/(xi-1)}^m

But how to proceed after this ? Hints will do..

Please help !

Thanks in advance...

First of all: what is meant by $\cot^{-1} x?$ (I know it, but do you?) Try to set $y = \cot^{-1} x$ and see where that gets you.

 

[Saitama]

Hello sankalp!

Look at $xi+1$. Can you convert it to e^{i*something}? :)

 

[sankalpmittal]

First of all: what is meant by $\cot^{-1} x?$ (I know it, but do you?) Try to set $y = \cot^{-1} x$ and see where that gets you.

I know what it is. Its an inverse trigonometric function with range 0 to pi, boundaries exclusive. Are you insisting to substitute y with cot inverse x ?

Hello sankalp!

Look at $xi+1$. Can you convert it to e^{i*something}? :)

Yeah. I got the answer. Thanks.

Rays approach also worked.

Thanks..

The question you asked in one post was from present fiitjee aits. Are you giving a second try for jee ?