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Problem
Problem
Prove that :
tan
−
1
(
e
i
θ
)
=
n
π
2
+
π
4
−
i
2
ln
tan
(
π
4
−
θ
2
)
,
where ’n’ is an integer.
\tan^{-1}(e^{i\theta}) = \frac{n\pi}{2} + \frac{\pi}{4} - \frac{i}{2} \ln \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right), \text{ where 'n' is an integer.}
tan
−
1
(
e
i
θ
)
=
2
nπ
+
4
π
−
2
i
ln
tan
(
4
π
−
2
θ
)
,
where ’n’ is an integer.
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