Complex numbers so complex (1 Viewer)

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I'm new to complex numbers and I am completely lost and worded explanations would be very beneficial.

1.
Is the set {-1,1} closed under addition, closed under multiplication ?
How about the sets {0} and {-1} ?

2.
Find (1 - i)² and hence find (1 - i)²⁰⁰⁸

3.
What is the difference in answering a question (e.g √3 + i)²⁰⁰⁸ in polar form and Cartesian form ?

4.
What's the deal with 2kπ in:
Arg(zw) = Arg(z) + Arg(w) + 2kπ
Arg(z/w) = Arg(z) - Arg(w) + 2kπ
Arg(zⁿ) = nArg(z) + 2kπ

5.
Is there a proof for e^iθ = cos θ + isinθ or e^iθ = e^i(θ+2kπ) ?

6.
Image Below

 

[me121]

3.
What is the difference in answering a question (e.g √3 + i)²⁰⁰⁸ in polar form and Cartesian form ?

An answer in Cartesian form will look like a + bi. however in polar it would look like r.cis(theta) = r.cos(theta) + ri.sin(theta).

Complex number's go on the complex name, so its just like representing a point on a number plane, you can give its coordinates by x,y (Cartesian) or r,theta (polar).

 

[Trebla]

I'm new to complex numbers and I am completely lost and worded explanations would be very beneficial.

1.
Is the set {-1,1} closed under addition, closed under multiplication ?
How about the sets {0} and {-1} ?

2.
Find (1 - i)² and hence find (1 - i)²⁰⁰⁸

3.
What is the difference in answering a question (e.g √3 + i)²⁰⁰⁸ in polar form and Cartesian form ?

4.
What's the deal with 2kπ in:
Arg(zw) = Arg(z) + Arg(w) + 2kπ
Arg(z/w) = Arg(z) - Arg(w) + 2kπ
Arg(zⁿ) = nArg(z) + 2kπ

5.
Is there a proof for e^iθ = cos θ + isinθ or e^iθ = e^i(θ+2kπ) ?

6.
Image Below

1) I'm not quite sure about this, but I think when you add -1 and 1 together you get 0 which is a number that is not within the set {-1,1} thus it is open in addition (I think). However when you multiply 1 and -1 however many times together you get - 1 or 1 which is a number within the set {-1,1}, so it is closed under multiplication (I think). From that, the set {0} is closed under both addition and multiplication because the only number within that set is 0, so when multiply 0 with 0 however many times, you still get zero, same with addition. For {-1}, its open in addition because when you add -1 however many times, the result is not always -1 so it is not within the set {-1}. With multiplication it is open because when you multiply -1 however many times, you get either 1 or -1 hence it is not always within the set {-1}. (I think)

2) (1 - i)² = 1 - 2i + i² = - 2i (since i² = -1)
Thus:
(1 - i)²⁰⁰⁸
= (-2i)¹⁰⁰⁴
= (-2)¹⁰⁰⁴.i¹⁰⁰⁴
= 2¹⁰⁰⁴.(i⁴)²⁵¹
= 2¹⁰⁰⁴ (since i⁴ = (i²)² = (-1)² = 1)

3) Cartesian form is in x + iy so you find the complex coordinates (x,y) on the Argand diagram. Polar form is r(cos θ + isinθ), where r is the modulus and θ is the argument so you draw a vector of length r from the positive real (x) axis and rotate it anti-clockwise by θ to find your complex coordinates. So it's really just two ways of finding the same point.

4) The addition of 2kπ (k is an integer), generalises the argument over all real numbers rather than the principal argument between - π and π inclusive, so: cos x + isin x = cos (x + 2π) + isin (x + 2π) = cos (x + 4π) + isin (x + 4π) = cos (x + 6π) + isin (x + 6π) and so on since the trig functions give the same value after adding 2π to the angle.

5) Euler's formula e^iθ = cos θ + isin θ can be proved using the Taylor series of exponentials and trigonometric functions which you'll cover later on but here are some other proofs below if you're interested: http://en.wikipedia.org/wiki/Euler's_formula#Proofs

As for proving e^iθ = e^{i(θ + 2kπ}), it's intuitive because cos θ + isin θ = cos (θ + 2π) + isin (θ + 2π) = cos (θ + 4π) + isin (θ + 4π) = cos (θ + 6π) + isin (θ + 6π) and so on...
Hence: cos θ + isin θ = cos (θ + 2kπ) + isin (θ + 2kπ) for integer k
=> e^iθ = e^{i(θ + 2kπ})

6) A = B, A =/= C, A =/= D, B =/= C, B =/= D, C=/D, B is contained within A, D is contained within A and B and the rest are not contained within each other (unless I've missed some)

Remember U = union (combine sets) and ∩ means intersection (find where they are both common)
AUC = {1,1,1,2,2,3,5,6,7,8,8,8}
A∩C = {1,2}
AUD = {1,1,2,2,5,5,8,8,8}
A∩D = {2,5}
CUD = {1,2,2,3,5,6,7}
C∩D = {2}

Hope that helps.

Yay 1000th post!

 

[Affinity]

{-1,1} is closed under multiplication but not addition,
{0} is closed under multiplication and addition
{-1} is not closed under either operation (you don't say open)

Yeah.. the 2kpi is there to get the principal argument, for example, if you take arguments to be between -pi to pi, then -1+i has argument 3pi/4 but
you don't have -pi/2 = Arg((-1+i)(-1+i)) =/= 2Arg(-1+i). ofcourse this problem could be avoided if we consider arg to be multivalued.. in the sense that 0,2pi,4pi etc are all the same.

euler's formula.. hmm.. this is more of a definition than theorem... because you do not have a definition for the complex exponential (unless you define e^x to be the power series sum which is equivalent). e^{ix} = cos(x) + isin(x) because it's the only definition which "makes sense"(compatible with what we know about e^x like e^a * e^b = e^{a+b})