Proving an identity in complex analysis.?

Question

I need lots of help in proving this identity.

For a complex number z, prove:

z=tan[(1/ i)(log((1+iz)/(1-iz))^(1/2))]

Answer 1

As was already said, this is not an identity. Still, we can simplify the expression on the right Let u = (i z - 1)/(i z + 1) and w = sqr(u). We want to simplify tan[ (log w)/ i ].

Using the basic exp( i theta) = cos(theta) + i sin(theta) and doing a little simplification we can find exponential formulas for sin, cos and then tan. We get

tan(theta) = - i * [exp(i theta) - exp( - i theta) ] / [exp(i theta) + exp( - i theta) ].

Let theta = log sqr (u) and use the fact that exp log sqr(u) = sqr(u) to see that the expression on the right simplifies to

- i * [sqr(u) - 1/sqr(u)] / [sqr(u) + 1/sqr(u)].

Then multiply top and bottom by sqr(u) to get

- i * ( u - 1 )/ (u + 1).

Now substitute for u in terms of z and simplify to get: the corrected answer which is 1/z.

Answer 2

Just expand in power series Arctan z on one side and
1/2i (log(1+iz) - log ( 1-iz) ). This makes sense for |z|<1.

Answer 3

take z = -2i and you see that youi cant prove this identity.
this is not an identity z is not equal for every z to the expression on the right side of the - sign