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### 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.
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```
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/
```
Associated Topics:
High School Imaginary/Complex Numbers
High School Trigonometry

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