Math 101 Revision - Lecture 12

[fardell24]

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|z| = √z z


Argand Diagram
Polar form of a complex number in polar coordinates (1, θ)
We have x = rcosθ
y = rsinθ
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note x2 + y 2 = r2cosθ + r2sinθ = r2 = zz
z = r(cosθ + isinθ)
Polar form of z

De Moivres Theorem
Note zn = rn (cosθ + isinθ)n , How do we calculate (cosθ + isinθ)n ?

Theorem (De Moivre) If n is a rational number then (cosθ + isinθ)n = cos(nθ) + isin(nθ)
Proof Exercise
Cos(A+B) = cosAcosB – SinAsinB
Sin(A+B) = SinAcosB + SinBcosA
e.g. Calculate: (i) (1 + i)4 (ii) (√3/2 + I 1/2)3
Solutions
(i) Polar form for 1 + I : -
Now |1 + i| = √(i2 + i2)
= √2
∴ (1 + i)/√2 = cosθ + isinθ

i.e. cosθ = 1/√2
} ∴ θ = π/4
= 45°
sinθ = 1/√2

∴ √2(cos(π/4) + isin(π/4)) – the polar form.
Then (1 + i)4 = (√2)4 x (cos(π/4) + isin(π/4))4
= (√2)4 x (cosπ + isinπ)
= 4(cosπ + isinπ)
= -4, as cosπ = -1 sinπ = 0

(ii) Polar form of √3/2 + i/2
|√3/2 + i/2| = √((√3/2)2 + (i/2)2)
= √(3/4 + 1/4)
= √1
= 1
(√3/2 + i/2) = cosθ + isinθ
With cos θ = √3/2
}
sin θ = ½

Polar form
√3/2 + i/2 = cos(π/6) + isin(π/6)
∴ (√3/2 + i/2)3 = (cos(π/6) + isin(π/6))3
= cos(π/2) + isin(π/2)
= i, cos(π/2) = 0, sin(π/2) = 1

Functions
f: A  B, sets A, B
f: a| b, f(a) = b, a € A, b € B
g: B  C
g: b <--> c, g(b) = c, b € B , c € C
g0 f(a) = g(f(a))

Injective (1 – 1, one to one)
For function f:A -> B
F: a|B
Given any b € B there is at most one a € A each that f(a) = b, the function is called injective
Surjective (onto)
A function of f:AB is called surjective if for every b € B there is at least one a € A such that f(a) = b.

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R – valued functions on |R



http://cubeupload.com/im/Brenorenz/Rvaluedfunction1.png


http://cubeupload.com/im/Brenorenz/Rvaluedfunction2.png

f: |R  |R
f(x) = - 1
Not injective and not surjective as a function from |R +-> |R

What about f:|R  {y € |R : u ≥ 0}
f(x) = x2


http://cubeupload.com/im/Brenorenz/Rvaluedfunction3.png

Not injective (two x’s for every y)
Is surjective.

What about f: {x € |R: x ≥ 0}  { y € |R: y ≥ 0
f(x) = x2



http://cubeupload.com/im/Brenorenz/Rvaluedfunction4.png

Both surjective and injective.

Surjective and injective is bijective.