# Thread: [SOLVED] solve cosh(x)= i

1. ## [SOLVED] solve cosh(x)= i

any help will be appreciated....thank you

best i got out of this was x=ln{1+ or - 2^(1/2)}i ...but unsure if that is correct or even a valid expression...thank you for any help ...peace

2. $\displaystyle \begin{array}{l} \cosh (z) = \cosh (x)\cos (y) + i\sinh (x)\sin (y) = i \\ z = \sinh ^{ - 1} (1) + \left( {\frac{{1}}{2}\pi } \right)i \\ \end{array}$

3. Originally Posted by crb
any help will be appreciated....thank you

best i got out of this was x=ln{1+ or - 2^(1/2)}i ...but unsure if that is correct or even a valid expression...thank you for any help ...peace
Initial equation
$\displaystyle cosh(x)=i$

Take inverse
$\displaystyle x=arcosh(i)$

Identity
$\displaystyle x=ln(i+\sqrt{i-1}\sqrt{i+1})$

Multiply Square Roots
$\displaystyle x=ln(i+\sqrt{(i-1)(i+1)})$

FOIL
$\displaystyle x=ln(i+\sqrt{i^2-1})$

Simplify
$\displaystyle x=ln(i+\sqrt{-2})$

--------------------
VERIFICATION:

Initial Equation
$\displaystyle cosh(x)=i$

Identity
$\displaystyle \frac{e^x}{2} + \frac{e^{-x}}{2}=i$

Rewrite
$\displaystyle \frac{e^x}{2} + \frac{1}{2e^x}=i$

$\displaystyle \frac{e^{ln(i+\sqrt{-2})}}{2} + \frac{1}{2e^{ln(i+\sqrt{-2})}}=i$

Simplify
$\displaystyle \frac{i+\sqrt{-2}}{2} + \frac{1}{2(i+\sqrt{-2})}=i$

Multiply by 2 and subtract i from both sides
$\displaystyle \sqrt{-2} + \frac{1}{i+\sqrt{-2}}=i$

Common denominator
$\displaystyle \frac{i\sqrt{-2}+\sqrt{-2}\sqrt{-2}}{i+\sqrt{-2}} + \frac{1}{i+\sqrt{-2}}=i$

Simplify
$\displaystyle \frac{i^2\sqrt{2}-\sqrt{2}^2+1}{i+\sqrt{-2}}=i$

Simplify
$\displaystyle \frac{-\sqrt{2}-1}{i+\sqrt{-2}}=i$

Multiply by the Denominator
$\displaystyle -\sqrt{2}-1=i(i+\sqrt{-2})$

Distribute
$\displaystyle -\sqrt{2}-1=i^2+i\sqrt{-2}$

Simplify
$\displaystyle -\sqrt{2}-1=i^2+i^2\sqrt{2}$

Simplify
$\displaystyle -\sqrt{2}-1=-1-\sqrt{2}$

Both sides are equal, thus the answer is correct.

4. thanks for replies...appreciate it......thank you

5. Originally Posted by angel.white
Initial equation
$\displaystyle cosh(x)=i$

Take inverse
$\displaystyle x=arcosh(i)$

...
I did not check your answer. But this is a questionable step. The cosh function defined on $\displaystyle \mathbb{C}$ does not have an inverse. The way Plato has it is how I would do it.