Trigonometry and Complex Numbers
Date: 08/05/98 at 12:33:41 From: Eric Wnag Subject: Trigonometry, complex numbers Simplify (sqrt 3 - i)^7 into the form a + bi.
Date: 08/05/98 at 13:06:24 From: Doctor Jaffee Subject: Re: Trigonometry, complex numbers Hi Eric, The trick to solving this problem is putting the number into complex polar form, using DeMoivre's Theorem, then transforming the answer back into rectangular form (a + bi). Here is my suggestion. Locate the point sqrt(3) - i on the complex coordinate plane, then draw a segment from the point to the origin and draw another segment from the point perpendicular to the positive x-axis. Using the Pythagorean Theorem you can determine that the hypotenuse of the triangle you have drawn has length 2. Furthermore, the angle between the hypotenuse and the positive x-axis is 30 degrees in a clockwise direction, so we'll call it -30 degrees. Now, polar coordinate form looks like r(cos T + i sin T) where r is the distance from the point to the origin (in this case, 2), and T is the angle that the segment from the point to the origin makes with the positive x-axis. (in this case, -30 degrees). So, our expression in complex polar form is: [2(cos(-30) + i sin(-30)]^7 According to DeMoivre's Theorem: [2(cos(-30) + i sin(-30)]^7 = 2^7 [cos((-30)(7)) + i sin((-30)(7))] We now can evaluate 2^7 = 128 and (-30)(7) = -210. The result is 128(cos(-210) + i sin(-210)). Note cos(-210) = -sqrt(3)/2 and sin(-210) = 1/2. Therefore, the final answer is -64*sqrt(3) + 64i. I hope this has helped. - Doctor Jaffee, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum