Trigonometry: Getting Started - De Moivre's Theorem
This is a free lesson from our course in Trigonometry
 
   
This lesson content walks you through the De Moivre's Theorem. The trigonometric and exponential formulation is explained introducing the complex number definition in standard form. The simplification division of complex numbers is performed with the use of exponential forms. The formula is important because it connects complex number and trigonometry. The special feature is that you'll find all this with video explanation by instructor in own handwriting, some examples and practice questions with solution. According to the theorem; a complex number raised to a given positive integral power is equal to the modulus of the number raised to the power multiplied by the amplitude times the given power. (More text below video...)
<h2> Trigonometry - De Moivre Theorem</h2> <p> De Moivers Theorem, video, complex number, trigonometry help, example, solution, cosine, sine, positive integral power, modulus, amplitude, practice questions, quizzes</p> <p> a complex number raised to a given positive integral power is equal to the modulus of the number raised to the power multiplied by the amplitude times the given power.</p>
Other useful lessons:
Using De Moivre's Theorem
Roots of Complex Numbers

(Continued from above)For example, if z = r (cos  + i sin ) is a complex number, then
              zn = rn[cos (n) + sin (n)],
where n is an integer.
The expression "cos + i sin " is sometimes abbreviated to "cis ".

Problems involving powers of complex numbers can be solved using binomial expansion, but applying De Moivre's theorem is usually more direct.
E.g. Write (3 + i)7 in the form x + iy.
Solution : First determine the radius,
r = |3 + i|
r = (x2 + y2)
r = (3 + 1)
r = 2

Since cos = 3/2 and sin = , must be in the first quadrant and = 30. Therefore,
(3 + i)7 = [2(cos 30 + i sin 30)]7
Using De-Moivre's theorem,
= [27(cos 7(30) + i sin 7 (30))]
= 128(cos 210 + i sin 210)
= 128 (-(3/2)-(1/2)i)
= -643 - 64i

De Moivre's theorem can be extended to roots of complex numbers yielding the nth root theorem.

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