Complex Numbers Tips: Find Mod/Arg, Express in Cartesian

In summary, this person is trying to solve a complex equation that involves the modulus and argument of a complex number. They first find the modulus and argument of the complex number, and then use De Moivre's Theorem to find the complex exponential. They use this information to solve for the zbar of the complex number. They also try to solve for cos(π/12), but get stuck on how to get the exact value.
  • #1
erpoi
6
0
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.
 
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  • #2
[itex] \sin\frac{-\pi}{3}=-\sin\frac{\pi}{3} [/itex]

[itex] w=|w|e^{i\theta} [/itex]
 
  • #3
dextercioby said:
[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 don't understand your notation in the second part. Would you be able to explain further? Thanks!
 
  • #4
I thought the complex exponential was the "polar form".
 
  • #5
exp(i*theta)=cis(theta). He may not know the exponential notation.
 
  • #6
"(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
Dick said:
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
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
Eulers Formula:

[tex]e^{ix} = \cos x + i \sin x[/tex]. This let's us write complexs numbers in better ways. O and, De Moirves Theorem is an easy proof with this baby.
 
  • #10
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
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
Erm, Halls, you're being obtuse for no good reason:

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

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

1. What is the modulus of a complex number?

The modulus of a complex number is the distance from the origin to the point representing the complex number on the complex plane. It is calculated by taking the square root of the sum of the squares of the real and imaginary parts of the complex number.

2. How do you find the argument of a complex number?

The argument of a complex number is the angle between the positive real axis and the vector representing the complex number on the complex plane. It is calculated by taking the inverse tangent of the imaginary part divided by the real part.

3. Can you express a complex number in Cartesian form?

Yes, a complex number can be expressed in Cartesian form as a combination of its real and imaginary parts, written as a + bi, where a is the real part and bi is the imaginary part.

4. What is the importance of finding the modulus and argument of a complex number?

The modulus and argument of a complex number provide important information about the number's magnitude and direction in the complex plane. They are useful in performing operations on complex numbers, such as addition, subtraction, multiplication, and division.

5. How do you convert a complex number from polar form to Cartesian form?

To convert a complex number from polar form (r, θ) to Cartesian form (a + bi), you can use the trigonometric identities: a = r cos θ and b = r sin θ. Simply plug in the values for r and θ to find the real and imaginary parts of the complex number.

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