- #1

- 187

- 0

tanh (iu) = i tan u

however i do not get the correct relationship, is this an error in the question perhaps

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- Thread starter thenewbosco
- Start date

- #1

- 187

- 0

tanh (iu) = i tan u

however i do not get the correct relationship, is this an error in the question perhaps

- #2

- 121

- 1

The question is correct. Remember i^2 = -1.

- #3

- 187

- 0

what i have done is:

for the left side:

[tex]tanh u = \frac{sinh u}{cosh u} = \frac{e^{iu}-e^{-iu}}{e^{iu}+e^{-iu}}[/tex]

but then right side

[tex] i tan u = i\frac{ \frac{e^u-e^{-u}}{2i} } \frac{ e^{iu}+e^{-iu} } {2} }=

\frac{e^{u}-e^{-u}}{e^{iu}+e^{-iu}}[/tex]

what have i done wrong here?

for the left side:

[tex]tanh u = \frac{sinh u}{cosh u} = \frac{e^{iu}-e^{-iu}}{e^{iu}+e^{-iu}}[/tex]

but then right side

[tex] i tan u = i\frac{ \frac{e^u-e^{-u}}{2i} } \frac{ e^{iu}+e^{-iu} } {2} }=

\frac{e^{u}-e^{-u}}{e^{iu}+e^{-iu}}[/tex]

what have i done wrong here?

Last edited:

- #4

- 187

- 0

the second part of the second equation there should be e^(iu)+e^(-iu)/2

- #5

- 121

- 1

[tex] tan u = i(e^{-iu} - e^{iu}) / (e^{-iu} + e^{iu}) [/tex]

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