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Complex Numbers tips

  • Thread starter erpoi
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  • #1
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I was wondering if someone can check my solutions and perhaps give me a faster more logical way of working through this question. Thanks.

Homework Statement



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)


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.
 

Answers and Replies

  • #2
dextercioby
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[itex] \sin\frac{-\pi}{3}=-\sin\frac{\pi}{3} [/itex]

[itex] w=|w|e^{i\theta} [/itex]
 
  • #3
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[itex] \sin\frac{-\pi}{3}=-\sin\frac{\pi}{3} [/itex]

[itex] w=|w|e^{i\theta} [/itex]
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
dextercioby
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I thought the complex exponential was the "polar form".
 
  • #5
Dick
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exp(i*theta)=cis(theta). He may not know the exponential notation.
 
  • #6
Dick
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"(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
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He may not know the exponential notation.
Yes I have never heard of it, does it make doing the question easier? If so, can you explain?
 
  • #8
HallsofIvy
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The "hence" part is due to this:
[tex]\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 [itex]z^5[/itex] so you can easily solve for [itex]\bar{z}^5[/itex].
 
  • #9
Gib Z
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Eulers Formula:

[tex]e^{ix} = \cos x + i \sin x[/tex]. This lets us write complexs numbers in better ways. O and, De Moirves Theorem is an easy proof with this baby.
 
  • #10
Gib Z
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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:
  • #11
Dick
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The nice thing about the exponential notation is that it turns an equation like [tex]cis(a)*cis(b)=cis(a+b)[/tex] into [tex]e^{i a} e^{i b}=e^{i (a+b)}[/tex]. 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
matt grime
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Erm, Halls, you're being obtuse for no good reason:

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

* is the same as the over bar.
 
  • #13
HallsofIvy
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Well, that happens occaissionally! Perhaps more often that I would like!
 

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