# How to take the real part of this function

1. Nov 9, 2008

### makethings

Hello. I have been trying to solve Stokes Second Problem modified with a stationary upper plate (with oscillating bottom plate with velocity u(0,t) = U cos nt). I found my solution for the velocity profile but I can't simplify it since my math knowledge is limited.

Now you don't need to know what Stokes Second Problem is, but I need some help trying to take the Real part of this solution.

$$u(y,t) = Re\{U\frac{e^{a(H-y)} e^{i[a(H-y) +nt]} - e^{-a(H-y)} e^{-i[a(H-y) +nt]}}{ e^{aH} e^{iaH} - e^{-aH} e^{-iaH}}\}$$

where U, a, H, are constants
the variable is y

appreciate any help. Thank you

2. Nov 9, 2008

### gabbagabbahey

The real part of any function can be found using;

$$\text{Re} [f(y)]=\frac{f(y)+\overline{f(y)}}{2}$$

where the overline denotes complex conjugation....does that help?

3. Nov 9, 2008

### makethings

You might have to point me to a tutorial on taking complex conjugates of a function as complicated as mine. I know I can break down $$e^{iz} = cosz + i sinz$$ and perhaps take the conjugate as that cos - i sin ?? But with a function with multiple terms containing i, what would I do?

4. Nov 9, 2008

### cup

Assuming that U, a, H are all real, it is very simple to take the complex conjugate: Just change the signs of all the i's.

5. Nov 9, 2008

### makethings

To clarify, is it changing the sign of the terms containing i, or i itself?

6. Nov 9, 2008

### cup

(In the following, z* denotes the complex conjugate of z.)
The first thing to do is to make the denominator real (you will see why this makes sense):

$$U \frac{ \left( e^{a(H-y)} e^{i[a(H-y) +nt]} - e^{-a(H-y)} e^{-i[a(H-y) +nt]} \right) \left( e^{aH} e^{iaH} - e^{-aH} e^{-iaH} \right)^{*} } { \left( e^{aH} e^{iaH} - e^{-aH} e^{-iaH} \right) \left( e^{aH} e^{iaH} - e^{-aH} e^{-iaH} \right)^{*} } =$$

$$U \frac{ \left( e^{a(H-y)} e^{i[a(H-y) +nt]} - e^{-a(H-y)} e^{-i[a(H-y) +nt]} \right) \left( e^{aH} e^{-iaH} - e^{-aH} e^{iaH} \right) } { \left( e^{aH} e^{iaH} - e^{-aH} e^{-iaH} \right) \left( e^{aH} e^{-iaH} - e^{-aH} e^{iaH} \right) } =$$

$$U \frac{ e^{a(2H-y)} e^{i[-ay+nt]} - e^{-ay} e^{i[a(2H-y)+nt]} - e^{ay} e^{-i[a(2H-y)+nt]} + e^{-a(2H-y)} e^{-i[-ay+nt]} } { e^{2aH} - e^{i2aH} - e^{-i2aH} + e^{-2aH} } =$$

$$U \frac{ e^{a(2H-y)} e^{i[-ay+nt]} - e^{-ay} e^{i[a(2H-y)+nt]} - e^{ay} e^{-i[a(2H-y)+nt]} + e^{-a(2H-y)} e^{-i[-ay+nt]} } { 2\cosh(2aH) - 2\cos(2aH) }$$

The real part of this expression is now easily seen (as described above) to be

$$U \frac{ e^{a(2H-y)} \cos[-ay+nt] - e^{-ay} \cos[a(2H-y)+nt] - e^{ay} \cos[a(2H-y)+nt] + e^{-a(2H-y)} \cos[-ay+nt] } { \cosh(2aH) - \cos(2aH) } =$$

$$2U \frac{ \cosh(a(2H-y)) \cos[-ay+nt] - \cosh(ay) \cos[a(2H-y)+nt] } { \cosh(2aH) - \cos(2aH) }$$

...I think. You should check it carefully.

Last edited: Nov 9, 2008
7. Nov 9, 2008

### cup

Example:

If

$$z = \frac{a-bi}{c+id}e^{-ig}$$

then

$$z^* = \frac{a+bi}{c-id}e^{+ig}$$

8. Nov 9, 2008

### makethings

Thanks..this math is tedious. I was working on it at the same time you were and we have ever so slightly different answers ( I got some - where you have + and I have a sinh appear).. now I got to nit pick and see where you or I went wrong. ugh.. lol Thanks though for doing that grunt work. appreciate it very much.

9. Nov 10, 2008

### cup

I find that it's easier to do these kinds of calculations on a computer, using Tex.
As long as you use a "vertical" coding style (one term per line only), it won't get messy.

The ability to copy and paste really helps.