# Multiplication z 1 z2 r 1 r 2 cis ϑ 1 ϑ 2 division

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Multiplication: z 1 × z 2 = r 1 × r 2 cis ϑ 1 + ϑ 2 Division: z 1 z 2 = r 1 r 2 cis ϑ 1 - ϑ 2 Example . Euler’s and De Moivre’s theorem These two theorems state the relationship between the trigonometric functions and the complex exponential function. This allows us to convert between Cartesian and Polar forms. Euler’s Theorem e i x = cos x + isin x De Moivre’s theorem z n = r ( cos x + isin x ) n = r n cos ( nx ) + isin ( nx ) De Moivre’s theorem can be derived from Euler’s through the exponential law for integer powers. ( e i x ) n = e i xn = z n 20
ALGEBRA Complex numbers 1 De Moivre’s theorem: proof by induction Having seen the method of induction, we will now apply it to De Moivre’s theorem. Prove: z n = cos ( x ) isin ( x ) n = cos ( nx ) + isin ( nx ) . 1. Show true for n = 1 cos ( x ) isin ( x ) 1 = cos ( 1 x ) + isin ( 1 x ) cos ( x ) isin ( x ) = cos ( x ) + isin ( 1 x ) is true for n = 1 2. Assume true for n = k cos ( x ) + isin ( x ) k = cos ( kx ) + isin ( kx ) 3. Prove true for n = k + 1 cos ( x ) + isin ( x ) k + 1 = cos ( k + 1 ) x + isin ( k + 1 ) x = cos ( x ) + isin ( x ) 1 cos ( x ) + isin ( x ) k Inductive step: use assumption about n = k = cos ( x ) + isin ( x ) cos ( kx ) + isin ( kx ) Remember i 2 = - 1 = cos ( x ) cos ( kx ) + cos ( x ) isin ( kx ) + isin ( x ) cos ( kx ) - sin ( x ) sin ( kx ) = cos ( x ) cos ( kx ) - sin ( x ) sin ( kx ) + i cos ( x ) sin ( kx ) + sin ( x ) cos ( kx ) Use of the double / half angle formulae = cos ϑ + k ϑ + isin ϑ + k ϑ = cos ( k + 1 ) ϑ + isin ( k + 1 ) ϑ is required result and form. Hence, by assuming n = k true, n = k + 1 is true. Since the statement is true for n = 1, it is true for all n Z + . Example . 21
ALGEBRA Complex numbers 1.5.3 Nth roots of a complex number n th root of a complex number z is a number ω such that ω n = z . To find n th root of a complex number (in polar form, z = r cis ( ϑ ) ) you need to use the following formula: z k + 1 = n p r × cis ϑ n + 2 k π n for k = 0,1,2,..., n - 1. Finding complex roots Find 3 roots of z 3 = 4 + 4 p 3i and draw them on the complex plane. 1. Rewrite the complex number in polar form. r = ¨ 4 2 + 4 p 3 2 = p 64 = 8 ϑ = arctan p 3 = π 3 z 3 = 8cis π 3 2. Insert values into formula. z 1 = 3 p 8 × cis π 3 × 3 = 2cis π 9 z 2 = 3 p 8 × cis π 3 × 3 + 2 × 1 × π 3 = 2cis 7 π 9 z 3 = 3 p 8 × cis π 3 × 3 + 2 × 2 × π 3 = 2cis 13 π 9 3. Draw the roots on the complex plane, they should be equally spaced out with the same length. z 1 z 2 z 3 22
ALGEBRA System of linear equations: Unique, infinite and no solutions 1 1.6 System of linear equations: Unique, infinite and no solutions Solving a system of two linear equations should be familiar to most of you. There are several methods of solving it, including substitution and subtraction of equations from each other. However, sometimes there can be three equations with three unknowns or even two equations with three unknowns. It is important to identify when those equations have unique, infinite amount or no solutions at all. The easiest way to do it is to solve the simultaneous equations. It is possible to think of a system of linear equations geometrically, where the solution is at the intersection of lines or planes. Thus the intersection can be a point, a line or a plane. Here is how to identify the amount of solutions that the system of equations has: Unique solution there is only one set of variables that satisfy all equations. Intersection is a point.