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Polar Form of Complex Numbers; DeMoivre's Theorem

Polar Form of Complex Numbers , DeMoivre's Theorem

History of Complex Numbers : Historically, complex numbers came about when trying to solve
quadratic equations such as

x2 + 2x + 2 = 0:

In general, when trying to solve the quadratic equation ax2 + bx + c = 0, we use the quadratic
formula . Applying this formula to our quadratic equation yields the two solutions

As in this example, the quadratic formula may yield solutions involving the square root of a negative
number. As such, we introduce the "complex number system " with the properties as the real
numbers with the added property that , so that

i2 = -1.

Some felt that no meaning could be given to such expressions , which were therefore termed "imag-
inary".

Definition: Complex numbers are represented by expressions of the form

z = a + bi

where a = Rez is the real part of z and b = Imz is called the imaginary part of z. Graphically, we
graph a complex number in the complex plane with a real axis (horizontal) and an imaginary axis
(vertical).

Exercises: Sketch the following complex numbers in the complex plane:

Definition: The modulus (or absolute value) of the complex number z = a + bi is

Definition: A complex number z = a + bi has polar (or trigonometric) form

z = r(cos θ + i sin θ),

where is the modulus of z and θ is the argument of z satisfying tan θ = b/a.

Exercises: Write each of the following complex numbers in polar form

Multiplication and Division of Complex Numbers: If

and ,

then

and if ,

Proof: This follows immediately from the sum and difference formulas for sine and cosine.

Exercises:

1. Write and in polar form and then find

2. Write z1 = 1 - i and in polar form and then find

3. Let z = r(cos θ + i sin θ). Find formulas for

DeMoivre's Theorem: If z = r(cos θ + i sin θ) and n is any positive integer, then

Exercise: With = , find z3.

nth Roots of a Complex Number: Reversing DeMoivre's Theorem will allow us to find the
n-roots of a complex number . That is, given z = a + bi, find w such that wn = z. To do so, we
write

z = r(cos θ + i sin θ)

so that one such w is

Replacing θ with θ + 2kπ for k = 0,... , n - 1 give all of the distinct nth roots of z. That is, the n
distinct nth roots of z = r(cos θ + i sin θ) are given by

Exercises:

1. Find the cube roots of i

2. Find the 6th roots of 1.

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