SLIDE 1
Complex Unit Circle
Polar coordinates
x2 = 1 has two solutions: x ∈ {±1}. Real Imaginary
Complex Unit Circle Polar coordinates x 2 = 1 has two solutions: x - - PowerPoint PPT Presentation
Complex Unit Circle Polar coordinates x 2 = 1 has two solutions: x - - PowerPoint PPT Presentation
Complex Unit Circle Polar coordinates x 2 = 1 has two solutions: x { 1 } . Imaginary Real x 3 = 1 has three solutions: x { 1 , 0 . 5 0 . 866 i } . Imaginary Real x 4 = 1 has four solutions: x { 1 , i } . Imaginary
SLIDE 2
SLIDE 3
x3 = 1 has three solutions: x ∈ {1, −0.5 ± 0.866i}. Real Imaginary
SLIDE 4
x4 = 1 has four solutions: x ∈ {±1, ±i}. Real Imaginary
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x5 = 1 has five solutions: x ∈ {1, 0.309 ± 0.951i, −0.809 ± 0.588i}. Real Imaginary
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eix = cos(x) + i sin(x) therefore e2πi = 1
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eix = cos(x) + i sin(x) therefore e2πi = 1 Raising both sides to the kth power we get e(2πi)k = 1k = 1 for k = 0, 1, 2, 3, . . .
SLIDE 8
eix = cos(x) + i sin(x) therefore e2πi = 1 Raising both sides to the kth power we get e(2πi)k = 1k = 1 for k = 0, 1, 2, 3, . . . xn = 1 has the solution x = e(2πi)k/n for k = 0, 1, 2, 3, . . .
SLIDE 9
eix = cos(x) + i sin(x) therefore e2πi = 1 Raising both sides to the kth power we get e(2πi)k = 1k = 1 for k = 0, 1, 2, 3, . . . xn = 1 has the solution x = e(2πi)k/n for k = 0, 1, 2, 3, . . . e(2πi)0/3 = 1 e(2πi)1/3 = −0.5 + 0.866i e(2πi)2/3 = −0.5 − 0.866i
SLIDE 10
The [MODE] menu has the option to represent complex numbers in the form x = reθi where r is the radius (complex absolute value) and θ is the angle. Alternativly, the [MATH][CPX][→Rect] and [→Polar] menu items can be used.