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Complex Numbers in Polar Form; DeMoivre’s Theorem 6.5

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Presentation on theme: "Complex Numbers in Polar Form; DeMoivre’s Theorem 6.5"— Presentation transcript:

1 Complex Numbers in Polar Form; DeMoivre’s Theorem 6.5

2 The Complex Plane We know that a real number can be represented as a point on a number line. By contrast, a complex number z = a + bi is represented as a point (a, b) in a coordinate plane, shown below. The horizontal axis of the coordinate plane is called the real axis. The vertical axis is called the imaginary axis. The coordinate system is called the complex plane. Every complex number corresponds to a point in the complex plane and every point in the complex plane corresponds to a complex number. Imaginary axis Real axis a b z = a + bi

3 Text Example Plot in the complex plane:
a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i Solution We plot the complex number z = 3 + 4i the same way we plot (3, 4) in the rectangular coordinate system. We move three units to the right on the real axis and four units up parallel to the imaginary axis. -5 -4 -3 -2 1 2 3 4 5 z = 3 + 4i z = -1 – 2i The complex number z = -1 – 2i corresponds to the point (-1, -2) in the rectangular coordinate system. Plot the complex number by moving one unit to the left on the real axis and two units down parallel to the imaginary axis.

4 Text Example cont. Plot in the complex plane:
a. z = 3 + 4i b. z = -1 – 2i c. z = -3 d. z = -4i Solution Because z = -3 = i, this complex number corresponds to the point (-3, 0). We plot –3 by moving three units to the left on the real axis. -5 -4 -3 -2 1 2 3 4 5 z = 3 + 4i z = -3 Because z = -4i = 0 – 4i, this complex number corresponds to the point (0, -4). We plot the complex number by moving three units down on the imaginary axis. z = -4i z = -1 – 2i

5 The Absolute Value of a Complex Number
The absolute value of the complex number a + bi is

6 Example Determine the absolute value of z=2-4i Solution:

7 Polar Form of a Complex Number
The complex number a + bi is written in polar form as z = r (cos  + i sin  ) where a = r cos  , b = r sin  , and tan =b/a The value of r is called the modulus (plural: moduli) of the complex number z, and the angle  is called the argument of the complex number z, with 0 <  < 2

8 Text Example Plot z = -2 – 2i in the complex plane. Then write z in polar form. Solution The complex number z = -2 – 2i, graphed below, is in rectangular form a + bi, with a = -2 and b = -2. By definition, the polar form of z is r(cos  + i sin  ). We need to determine the value for r and the value for  , included in the figure below. Imaginary axis 2 è Real axis -2 2 r -2 z = -2 – 2i

9 Text Example cont. Solution
Since tan p4 = 1, we know that  lies in quadrant III. Thus,

10 Product of Two Complex Numbers in Polar Form
Let z1 = r1 (cos 1+ i sin  1) and z2 = r2 (cos  2 + i sin  2) be two complex numbers in polar form. Their product, z1z2, is z1z2 = r1 r2 (cos ( 1 +  2) + i sin ( 1 +  2)) To multiply two complex numbers, multiply moduli and add arguments.

11 z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º)
Text Example Find the product of the complex numbers. Leave the answer in polar form. z1 = 4(cos 50º + i sin 50º) z2 = 7(cos 100º + i sin 100º) Solution z1z2 = [4(cos 50º + i sin 50º)][7(cos 100º + i sin 100º)] Form the product of the given numbers. Multiply moduli and add arguments. = (4 · 7)[cos (50º + 100º) + i sin (50º + 100º)] = 28(cos 150º + i sin 150º) Simplify.

12 Quotient of Two Complex Numbers in Polar Form
Let z1 = r1 (cos 1 + i sin 1) and z2 = r2 (cos 2 + i sin 2) be two complex numbers in polar form. Their quotient, z1/z2, is To divide two complex numbers, divide moduli and subtract arguments.

13 DeMoivre’s Theorem Let z = r (cos  + i sin ) be a complex numbers in polar form. If n is a positive integer, z to the nth power, zn, is

14 Text Example Find [2 (cos 10º + i sin 10º)]6. Write the answer in rectangular form a + bi. Solution By DeMoivre’s Theorem, [2 (cos 10º + i sin 10º)]6 = 26[cos (6 · 10º) + i sin (6 · 10º)] Raise the modulus to the 6th power and multiply the argument by 6. = 64(cos 60º + i sin 60º) Simplify. Write the answer in rectangular form. Multiply and express the answer in a + bi form.

15 DeMoivre’s Theorem for Finding Complex Roots
Let =r(cos+isin) be a complex number in polar form. If 0,  has n distinct complex nth roots given by the formula

16 Example Find all the complex fourth roots of 81(cos60º+isin60º)
Solution:

17 Example cont. Find all the complex fourth roots of 81(cos60º+isin60º)
Solution:


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