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