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I want to show $\Re{(\cosh z)}) = \cosh(\Re z)\cdot \cos(\Im z)$.

I did the following:

\begin{align*} \cosh(\Re z)\cdot \cos(\Im z) &= \frac {1}{2}(e^{\Re z}+e^{-\Re z})\cdot \frac {1}{2}(e^{i\Im z}+e^{-i\Im z}) \\& = \frac {1}{4}(e^{\Re z}+e^{-\Re z})\cdot (e^{i\Im z}+e^{-i\Im z}) \\ &= \frac {1}{4}(e^{\Re z}+e^{-\Re z}) \cdot 2\\ & = \frac {1}{2}(e^{\Re z}+e^{-\Re z}) \\ &= \frac {1}{2}\Re(e^z+e^{-z}) \\ &= \Re(\cosh z) \end{align*}

Using: $\cosh(x) = \frac {1}{2}(e^x+e^{-x})$ and $\cos(x) = \frac {1}{2}(e^{ix}+e^{-ix})$.

I would be very glad if someone could tell me if this is correct!

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  • $\begingroup$ Delete the first equality for sure. Now as written, it looks like you assume the result and finagle around a bit. That's not what you've done. $\endgroup$ – Andres Mejia Aug 28 '18 at 22:42
  • $\begingroup$ @AndresMejia Is this better? $\endgroup$ – Zelda_CompSci Aug 28 '18 at 22:44
  • $\begingroup$ yeah, for sure. I edited the formatting a bit for readability. I hope you find that okay. $\endgroup$ – Andres Mejia Aug 28 '18 at 22:46
  • $\begingroup$ @AndresMejia Sure, thanks! Better formatting is always welcome. $\endgroup$ – Zelda_CompSci Aug 28 '18 at 22:47
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    $\begingroup$ I don't see how you got the third equality. $\endgroup$ – Andres Mejia Aug 28 '18 at 22:49
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As an alternative, but taking the path of OP so that his /her error can be seen: $$ \cosh\left[\operatorname{Re}(z)\right] \cdot \cos\left[\operatorname{Im}(z)\right] = \left(\frac{e^{\operatorname{Re}(z)} + e^{-\operatorname{Re}(z)}}{2}\right) \left(\frac{e^{i\operatorname{Im}(z)} + e^{-i\operatorname{Im}(z)}}{2}\right)$$ $$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)}}{4} $$ $$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) + i\operatorname{Im}(z)} + }{4} $$ $$ = \frac{e^{Re(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) - i\operatorname{Im}(z)]}}{4} $$ $$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) + i\operatorname{Im}(z)]}}{4} + \frac{e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) - i\operatorname{Im}(z)]}}{4} = \frac{\cosh {z}}{2} + \frac{\cosh {\overline {z}}}{2}$$ $$ = \frac{\cosh {z} + \cosh {\overline {z}}}{2} = \frac{2\operatorname{Re}{\left[\cosh{z}\right]}}{2} = \operatorname{Re}{\left[\cosh{z}\right]}$$

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Let $z=x+iy$, then we have that

$$\cosh z=\frac{e^z+e^{-z}}2=\frac12e^xe^{iy}+\frac12e^{-x}e^{-iy}=\frac12e^x(\cos y+i\sin y)+\frac12e^{-x}(\cos y-i\sin y) $$

$$ \implies \operatorname{Re}(\cosh z)=\frac12e^x\cos y+\frac12e^{-x}\cos y=\frac{e^x+e^{-x}}2\cos y=\cosh x \cdot\cos y$$

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  • $\begingroup$ This doesn't really answer the question in the body. $\endgroup$ – Andres Mejia Aug 28 '18 at 23:03
  • $\begingroup$ I should clarify that I did not downvote $\endgroup$ – Andres Mejia Aug 28 '18 at 23:11
  • $\begingroup$ @AndresMejia Yes you are right! I've just revised the derivation. $\endgroup$ – gimusi Aug 28 '18 at 23:14
  • $\begingroup$ Upvoted simply because I consider this the most direct and clearest proof. It seems some of the other answers are unnecessarily obfuscated because they wanted to mirror what the OP did. We shouldn't fall into this trap - it's never a given that the OP has found the simplest solution or the clearest way to present something. This solution gives some valuable tips - in a problem like this, encode the real and imaginary parts into single letter variables for clarity. Invoke the trigonometric form early to help split the real and complex parts. All these things will make future problems easier. $\endgroup$ – Deepak Aug 30 '18 at 10:22
  • $\begingroup$ @Deepak Thanks a lot for you kind appreciation! I must agree with you :P $\endgroup$ – gimusi Aug 30 '18 at 10:26
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Using the sum of angles formulae,

$\cosh(x+iy)=\cosh(x)\cosh(iy)+\sinh(x)\sinh(iy)$

but $ \cosh(iy)=\cos(y)$ and $\sinh(iy)=i\sin(y)$

so: $\cosh(x+iy)=\cosh(x)\cos(y) + i \sinh(x)\sin(y)$

QED

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