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Question : Prove that tan iz=-i tanh z ?
Answer :
A complex number z is written as
\[z=x+iy\]
where x and y are real numbers.
By the formula
\[tan (iz) = \frac{sin (iz)}{cos (iz)}\]
Now sin (iz) =i sinh z and cos (iz) = cosh z
\[tan (iz) = \frac{i sinh z}{cosh z} = i tanh z\]
\[tanh z = \frac{1}{i} tan (iz)\]
Thus we proved that tan iz=-i tanh z ?
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