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Complex Numbers, the Complex Plane & Demoivre’s Theorem Complex Numbers are numbers in the form of a bi where a and b are real numbers and i, the imaginary unit, is defined as follows: i 1 2 i 1 And the powers of i are as follows: i i 2 i 1 3 2 i i i 1 i i 4 i 1 1 The value of in, where n is any number can be found by dividing n by 4 and then dealing only with the remainder. Why? Examples: 1) i ? 18 18 4 4 with a remainder of 2 i i Then from the chart on the previous slide 18 2 i i 1 18 2) 2 i ? 27 4 6 with a remainder of 3 27 3 i i Then from the chart on the previous slide 27 i i i 27 3 In a complex number a bi a is the real part and bi is the imaginary part. When b=0, the complex number is a real number. When a0, and b0, as in 5+8i, the complex number is an imaginary number. When a=0, and b0, as in 5i, the complex number is a pure imaginary number. Lesson Overview 9-5A Lesson Overview 9-5B 5-Minute Check Lesson 9-6A The Complex Plane Imaginary z a bi Axis O Real Axis Let z a bi be a complex number. The magnitude or modulus of z, denoted by z As the distance from the origin to the point (x, y). z a b 2 2 is defined Imaginary Axis y z a bi |z| O x Real Axis r cos i sin is sometimes abbreviated as r cis z =-3 + 4i -3 Imaginary 4 Axis Real Axis z = -3 + 4i is in Quadrant II x = -3 and y=4 z =-3 + 4i 4 r 5 -3 Find the reference angle () by solving y tan x 4 tan 3 1 4 tan 3 53.13 z =-3 + 4i 4 r 5 -3 Find r: r 3 1 2 r2 2 Imaginary 4 Axis 3 -3 2 1 Real Axis 3 i Find the reference angle () by solving 3 1 2 y tan x 1 tan 3 1 1 tan 3 30 3 1 2 360 30 330 z r cos i sin z 2cos 330 i sin 330 Find the cosine of 330 and substitute the value. Find the sine of 330 and substitute the value. Distribute the r Write in standard (rectangular) form. 5 5 2 cos i sin 6 6 5 3 cos 6 2 3 1 2 i 2 2 5 1 sin 6 2 3 i Lesson Overview 9-7A Product Theorem Lesson Overview 9-7B Quotient Theorem 5-Minute Check Lesson 9-8A 5-Minute Check Lesson 9-8B Powers and Roots of Complex Numbers DeMoivre’s Theorem 81 81 3 i 2 2 What if you wanted to perform the operation below? 6 2 3i 4 Lesson Overview 9-8A Lesson Overview 9-8B Theorem Finding Complex Roots roots Find the complex fifth roots of The five complex roots are: for k = 0, 1, 2, 3,4 . 32 0i 2 2 32 0 r a b 2 tan 1 2 0 32 tan 1 0 322 32 0 32 cos 0 i sin 0 5 0 360k 0 360k 32 cos i sin 5 5 5 5 2 cos 72k i sin 72k 2 cos 72k i sin 72k k 0 2 cos 0 i sin 0 2 1 0i k 1 2 cos 72 i sin 72 .62 1.90i k 2 2 cos 144 i sin 144 1.62 1.18i k 3 2 cos 216 i sin 216 1.62 1.18i k 4 2 cos 288 i sin 288 .62 1.90i 2 The roots of a complex number a cyclical in nature. That is, when the points are plotted on a polar plane or a complex plane, the points are evenly spaced around the origin Complex Plane 4 2 5 -5 -2 -4 6 Polar plane 4 2 5 5 2 4 6 Polar plane 4 2 5 5 2 4 6 44 22 -5 5 55 -2 2 -4 4 To find the principle root, use DeMoivre’s theorem using rational exponents. That is, to find the principle pth root of a bi Raise it to the 1 power p Example Find 3 i First express You may assume it is the principle root you are seeking unless specifically stated otherwise. i 0i as a complex number in standard form. Then change to polar form b tan a 1 tan 0 r a 2 b2 1 tan 0 1 cos 90 i sin 90 1 r 1 90 1 cos 90 i sin 90 Since we are looking for the cube root, use DeMoivre’s Theorem and raise it to the 1 power 3 1 1 1 cos 90 i sin 90 3 3 1 3 1 cos 30 i sin 30 3 1 1 i 3 1 i 2 2 2 2 Example: Find the 4th root of 2 i 2 i 1 4 Change to polar form 5 cos 153.4 i sin 153.4 Apply DeMoivre’s Theorem 1 1 5 cos 153.4 i sin 153.4 4 4 1 4 1.22 cos 38.4 i sin 38.4 .96 .76i