Homework Help: Complex Numbers tips

1. Feb 20, 2007

erpoi

I was wondering if someone can check my solutions and perhaps give me a faster more logical way of working through this question. Thanks.

1. The problem statement, all variables and given/known data

If z = 1 + i*root3

i) Find the modulus and argument of z
ii) Express z^5 in Cartesian form a + ib where a and b are real
iii) Find z*zbar
iv) hence, find zbar to the power of negative 5

If w = root2 cis (π/4)

i) Find w/z in polar form.

ii) Express z and w in cartesian form and hence find w/z in cartesian form.

iii) Use answers from i) and ii) to deduce exact value for cos(π/12)

3. The attempt at a solution

i) Modulus is 2, Argument is π/3

ii) Using De Moivre's Theorem, 2^5cis(5*(π/3))
= 32cis(-pie/3)
Then changing to Cartesian form - 32 (cos(-π/3) + isin(-π/3))
= 32(0.5 + (root3/2)i)
= 16 + 16i*root3

iii) z = 1 + i*root3, zbar = 1 - i*root3
z*zbar = 4

iv) Now this part I don't understand the "hence", as in how am I meant to use previous results to get this answer?

I try - zbar = 2cis(-π/3) -> (Is it true that the "bar" of any complex number is just the same modulus and negative angle?)

Then zbar^-5 = Using DM Theorem, (1/32)cis(5π/3)
= (1/32)cis(-π/3)
Then convert to cartesian - (1/32)(cos(-π/3) + isin(-π/3))
= (1/32)(0.5 + iroot3/2)
= 1/64 + iroot3/64

If w = root2 cis (π/4)

i) Find w/z in polar form.
I found it in cartesian - (1+i)/(1 + root3 i) then realising, gives (1 + root3)/4 + (1-root3)/4 * i

But how do I change to Polar form? I am also stuck on how to get cos(pie/12) as exact value. Thanks.

2. Feb 20, 2007

dextercioby

$\sin\frac{-\pi}{3}=-\sin\frac{\pi}{3}$

$w=|w|e^{i\theta}$

3. Feb 20, 2007

erpoi

Hello I'm still very basic in this, and I dont understand your notation in the second part. Would you be able to explain further? Thanks!

4. Feb 20, 2007

dextercioby

I thought the complex exponential was the "polar form".

5. Feb 20, 2007

Dick

exp(i*theta)=cis(theta). He may not know the exponential notation.

6. Feb 20, 2007

Dick

"(Is it true that the "bar" of any complex number is just the same modulus and negative angle?)"

Sure. bar(cis(a))=cis(-a). It's also handy to convince yourself that cis(a)*cis(b)=cis(a+b) and cis(a)/cis(b)=cis(a-b). This is easy to see in the exponential notation. Now note pi/12=pi/3-pi/4. Does that help?

7. Feb 20, 2007

erpoi

Yes I have never heard of it, does it make doing the question easier? If so, can you explain?

8. Feb 20, 2007

HallsofIvy

The "hence" part is due to this:
$$\left(z\bar{z}\right)^5= z^5 \bar{z}^5[/itex] You know that the left side is 45= 1024 and you have already calculated $z^5$ so you can easily solve for $\bar{z}^5$. 9. Feb 20, 2007 Gib Z Eulers Formula: [tex]e^{ix} = \cos x + i \sin x$$. This lets us write complexs numbers in better ways. O and, De Moirves Theorem is an easy proof with this baby.

10. Feb 20, 2007

Gib Z

O and just incase you want to see how its proved, expand and simplify the taylor series of e^(ix) and see wat you get, or

let f(x)=cis(x)/e^(ix), find f'(x), once u prove its zero, just find one value and its obvious.

EDIT: O and iii is much easier with it i think, but i have to go, someone else will tell you.

Last edited: Feb 20, 2007
11. Feb 20, 2007

Dick

The nice thing about the exponential notation is that it turns an equation like $$cis(a)*cis(b)=cis(a+b)$$ into $$e^{i a} e^{i b}=e^{i (a+b)}$$. Written this way you don't have to remember it as a special rule for the cis function. It's a general property of exponentials. Because cis IS an exponential.

12. Feb 20, 2007

matt grime

Erm, Halls, you're being obtuse for no good reason:

(z*)^5 = (z^5)*

* is the same as the over bar.

13. Feb 20, 2007

HallsofIvy

Well, that happens occaissionally! Perhaps more often that I would like!