# Show that tan(i log a-ib/a+ib)=2ab/a^2-b^2

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### 1 Answer

We have to prove that: tan [i*log (a-ib/a+ib)] = 2ab/(a^2 - b^2)

Start with denoting a + ib in the form r*( cos x + i sin x)

Equate the real and complex components to get r*cos x = a and r*sin x = b

tan x = sin x / cos x = b/a

In the same notation a – ib = r( cos x – i*sin x)

The left hand side of what we have to prove is:

tan [i*log (a-ib/a+ib)]

=> tan [ i*log (r( cos x – i*sin x)/ r*( cos x + i sin x))]

we can write cos x – i*sin x as e^(-ix) and cos x + i*sin x as e^(ix)

=> tan[ i*log (e^(-ix)/e^(ix))]

=> tan[ i*log (e^(-2ix))]

=> tan[i (-2ix) log e]

log e = 1

=> tan [ -2*i^2*x]

=> tan( -2*-1*x)

=> tan 2x

expanding tan 2x

=> 2 tan x / 1 – (tan x)^2

=> [2*(b/a)] / [ 1 – (b/a)^2]

=> 2ab / a^2 + b^2

which is the right hand side.

**Therefore we prove that tan [i*log (a-ib/a+ib)] = 2ab/(a^2 - b^2)**