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06 complex exponentials contd, roots of unity

# 06 complex exponentials contd, roots of unity - 18.03...

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18.03 Lecture #6 Sept. 21, 2009: notes Four topics today: roots of complex numbers; calculus with complex numbers; trigonometry with complex numbers; and then a warmup for the “input/response” point of view on linear ODE. First topic: roots of complex numbers (Notes C4). To see that these can look a bit surprising, consider the calculation (1 / 2 + i/ 2) 2 = (1 2 / ( 2) 2 - i 2 / ( 2) 2 + i ((1 + 1) / ( 2) 2 = 0 + i (2 / 2) = i. This means that 1 / 2 + i/ 2 is a square root of i . This is easy enough to check ; the question is how you can figure it out in advance. Remember the polar representation of a complex number z = a + bi = re = r cos θ + ir sin θ. (Polar) The a + bi representation is perfect for addition of complex numbers, and the polar representation is perfect for multiplication: if w = se is a polar representation of another complex number, then z · w = ( re )( se ) = rse i ( θ + φ ) . (Polar Products) From this formula follows immediately (by induction on n ) z n = r n e inθ ( n any positive integer) (Polar powers) (In fact negative integers and zero work too.) Once we know how to do something, we also know how to undo it. From knowing how to take n th powers, we can deduce how to take n th roots. The n th roots of a complex number z (which are all distinct unless z = 0) are r 1 /n e ( θ +2 kπi ) /n ( n any positive integer, k = 1 , . . . n ) . (Polar roots) (If you like C++, then you’ll prefer to have k run from 0 to n - 1. That’s fine too.) This means that the n th roots of 1 are e (2 kπi ) /n = cos(2 kπ/n ) + i sin(2 kπ/n ) ( n any positive integer, k = 1 , . . . n ) . (Polar roots of 1)

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