# Integration, complex roots and radians

1. Mar 30, 2006

### shan

The first one, integration, I just want to check my answer.

$$\int \frac{1}{64} (\cos6\theta + 6\cos4\theta + 15\cos2\theta + 20) = \frac{1}{64} (\frac{\sin6\theta}{6} + \frac{6\sin4\theta}{4} + \frac{15\sin2\theta}{2} + 20\theta + c$$

I just wasn't sure if the integral of a constant wrt theta was constant*theta.

The complex roots question:
$$z^5 - i = (z-w1)(z-w2)(z-w3)(z-w4)(z-w5) = 0$$
Show that w1 + w2 + w3 + w4 + w5 = 0

From what our lecturer told us, $$-\prod_{1}$$ = 0 so the sum of the roots = 0. I don't really understand this though

The radians problem... is basically because I don't know how to express radian in decimals into pi radians. [As part of an answer for coshz = 2i] I want to express the argument of $$2 \pm \sqrt{5}$$ which would be $$\tan^{-1} \sqrt{5}/2$$.
In case you want to know, I found $$z = log(2 \pm i\sqrt{5}) = ln|2 \pm i\sqrt{5}| + i arg(2 \pm i\sqrt{5})$$
I just need help finishing up the question :)

Last edited: Mar 30, 2006
2. Mar 30, 2006

### dextercioby

Do you mean...?

$$\int 0 \ d\theta = \mbox{const}_{1} \cdot \theta + \mbox{const}_{2}$$

If so, then it's simply wrong, unless $\mbox{const}_{1}=0$ which doesn't make any sense to write in the first place...

For the second problem, couldn't u use a formula due to Viète...?

Daniel.

3. Mar 30, 2006

### AlphaNumeric

For the roots one, try writing i as $$i = e^{\frac{i\pi}{2}+2\pi ni}$$ and taking the fifth root of that. Then you can sum a geometric series to give you the answer.

4. Mar 30, 2006

### shan

oops sorry, I missed out a 20 in the integral so I'm asking if

$$\int 20 d\theta = 20\theta$$

I have never heard of this forumla?

And also, AlphaNumeric, could you explain why you suggested $$i = e^{\frac{i\pi}{2}+2\pi ni}$$??

5. Mar 30, 2006

### AlphaNumeric

$$z^{5} = i$$

You're looking for the 5 solutions of this equation, you're $$\omega_{1}$$ to $$\omega_{5}$$

They are clearly of modulus 1, and complex, so you'll have $$\omega_{n} = e^{it}$$ for some t you need to find. They are easy to find if you write i in for form I suggest.

$$z^{5} = e^{\frac{i\pi}{2}+2\pi n i}$$
Take 5th roots to get
$$z = e^{\frac{i\pi}{10}+\frac{2\pi n i}{5}$$

Usually express arguments in the range 0 to 2pi, so find the values of n which give you an argument in that range. n=0,1,2,3,4 will give you that, giving you roots

$$w_{1} = e^{\frac{i\pi}{10}}$$
$$w_{2} = e^{\frac{i\pi}{10}+\frac{2\pi i}{5}} = e^{\frac{5i\pi}{10}} = e^{\frac{i\pi}{2}} = i$$
etc

There's your 5 roots. Now you want to show they sum to zero. Well there's 5 of them and they are in geometric series, and you know how to sum those. The general formula is

$$S_{n} = a\frac{r^{n}-1}{r-1}$$

In this case $$r = e^{\frac{2\pi i}{5}}$$, so $$r^{5} = 1$$, and so $$S_{n} = 0$$.

Last edited: Mar 30, 2006
6. Mar 31, 2006

### shan

ok, thanks for your help alphanumeric, it's a little easier to see why it is true that way.

7. Mar 31, 2006

### AlphaNumeric

For a more graphical representation for why it's true, think about an Argand diagram. If you've the equation $$z^{n} = k$$ for any complex k, then the roots will be centred about the origin in a circle spread at an angle $$\frac{2\pi}{n}$$ radians apart from one another. You can see this from writing $$k = re^{i\theta}$$ (as I did for i) and repeating that method.

The average value of the n roots will be their sum divided by n. Geometrically, this is the centre of the circle, the origin. Therefore they sum to zero.

As an example, consider $$z^{n} = 16$$. Solutions are 2, -2, 2i, -2i. Plot them in the complex plane and you'll see they lie on a circle radius 2, with the circle centred on the origin.

I hope that helps you're understanding a bit

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