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Products of complex numbers in polar form

If z1=r1(cosθ1+isinθ1) and z2=r2(cosθ2+isinθ2), then the product of these numbers is given as:

z1z2=r1r2[cos(θ1+θ2)+isin(θ1+θ2)]z1z2=r1r2cis(θ1+θ2)

Notice that the product calls for multiplying the moduli and adding the angles.

Finding the product of two complex numbers in polar form

Find the product of z1z2, given z1=4(cos(80°)+isin(80°)) and z2=2(cos(145°)+isin(145°)).

Follow the formula

z1z2=42[cos(80°+145°)+isin(80°+145°)]z1z2=8[cos(225°)+isin(225°)]z1z2=8[cos(5π4)+isin(5π4)]z1z2=8[22+i(22)]z1z2=424i2
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Finding quotients of complex numbers in polar form

The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.

Quotients of complex numbers in polar form

If z1=r1(cosθ1+isinθ1) and z2=r2(cosθ2+isinθ2), then the quotient of these numbers is

z1z2=r1r2[cos(θ1θ2)+isin(θ1θ2)],z20z1z2=r1r2cis(θ1θ2),z20

Notice that the moduli are divided, and the angles are subtracted.

Given two complex numbers in polar form, find the quotient.

  1. Divide r1r2.
  2. Find θ1θ2.
  3. Substitute the results into the formula: z=r(cosθ+isinθ). Replace r with r1r2, and replace θ with θ1θ2.
  4. Calculate the new trigonometric expressions and multiply through by r.

Finding the quotient of two complex numbers

Find the quotient of z1=2(cos(213°)+isin(213°)) and z2=4(cos(33°)+isin(33°)).

Using the formula, we have

z1z2=24[cos(213°33°)+isin(213°33°)]z1z2=12[cos(180°)+isin(180°)]z1z2=12[1+0i]z1z2=12+0iz1z2=12
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Find the product and the quotient of z1=23(cos(150°)+isin(150°)) and z 2 = 2 ( cos ( 30° ) + i sin ( 30° ) ) .

z 1 z 2 = 4 3 ; z 1 z 2 = 3 2 + 3 2 i

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Finding powers of complex numbers in polar form

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem    . It states that, for a positive integer n , z n is found by raising the modulus to the n th power and multiplying the argument by n . It is the standard method used in modern mathematics.

De moivre’s theorem

If z = r ( cos θ + i sin θ ) is a complex number, then

z n = r n [ cos ( n θ ) + i sin ( n θ ) ] z n = r n cis ( n θ )

where n is a positive integer.

Evaluating an expression using de moivre’s theorem

Evaluate the expression ( 1 + i ) 5 using De Moivre’s Theorem.

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write ( 1 + i ) in polar form. Let us find r .

r = x 2 + y 2 r = ( 1 ) 2 + ( 1 ) 2 r = 2

Then we find θ . Using the formula tan θ = y x gives

tan θ = 1 1 tan θ = 1 θ = π 4

Use De Moivre’s Theorem to evaluate the expression.

( a + b i ) n = r n [ cos ( n θ ) + i sin ( n θ ) ] ( 1 + i ) 5 = ( 2 ) 5 [ cos ( 5 π 4 ) + i sin ( 5 π 4 ) ] ( 1 + i ) 5 = 4 2 [ cos ( 5 π 4 ) + i sin ( 5 π 4 ) ] ( 1 + i ) 5 = 4 2 [ 2 2 + i ( 2 2 ) ] ( 1 + i ) 5 = 4 4 i
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Finding roots of complex numbers in polar form

To find the n th root of a complex number in polar form, we use the n th Root Theorem or De Moivre’s Theorem    and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding n th roots of complex numbers in polar form.

The n Th root theorem

To find the n th root of a complex number in polar form, use the formula given as

z 1 n = r 1 n [ cos ( θ n + 2 k π n ) + i sin ( θ n + 2 k π n ) ]

where k = 0 , 1 , 2 , 3 , . . . , n 1. We add 2 k π n to θ n in order to obtain the periodic roots.

Practice Key Terms 4

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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