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Mathematics LibreTexts

21.2: Multiplication and division of complex numbers

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One interesting feature of the polar form of a complex number is that the multiplication and division are very easy to perform.

Proposition: Product and Quotient of Complex numbers

Let r1(cos(θ1)+isin(θ1)) and r2(cos(θ2)+isin(θ2)) be two complex numbers in polar form. Then, the product and quotient of these are given by

r1(cos(θ1)+isin(θ1))r2(cos(θ2)+isin(θ2))=r1r2(cos(θ1+θ2)+isin(θ1+θ2))

r1(cos(θ1)+isin(θ1))r2(cos(θ2)+isin(θ2))=r1r2(cos(θ1θ2)+isin(θ1θ2))

Proof

The proof uses the addition formulas for trigonometric functions sin(α+β) and cos(α+β) from proposition [PROP:trig-add-subt-formulas].

r1(cos(θ1)+isin(θ1))r2(cos(θ2)+isin(θ2))=r1r2(cos(θ1)cos(θ2)+icos(θ1)sin(θ2)+isin(θ1)cos(θ2)+i2sin(θ1)sin(θ2))=r1r2((cos(θ1)cos(θ2)sin(θ1)sin(θ2))+i(cos(θ1)sin(θ2)+sin(θ1)cos(θ2)))=r1r2(cos(θ1+θ2)+isin(θ1+θ2))

For the division formula, note, that the multiplication formula ??? gives

r2(cos(θ2)+isin(θ2))1r2(cos(θ2)+isin(θ2))=r21r2(cos(θ2θ2)+isin(θ2θ2))=1(cos0+isin0)=1(1+i0)=11r2(cos(θ2)+isin(θ2))=1r2(cos(θ2)+isin(θ2))

so that

r1(cos(θ1)+isin(θ1))r2(cos(θ2)+isin(θ2))=r1(cos(θ1)+isin(θ1))1r2(cos(θ2)+isin(θ2))=r1(cos(θ1)+isin(θ1))1r2(cos(θ2)+isin(θ2))=r1r2(cos(θ1θ2)+isin(θ1θ2))

Example 21.2.1

Multiply or divide the complex numbers, and write your answer in polar and standard form.

  1. 5(cos(11)+isin(11))8(cos(34)+isin(34))
  2. 3(cos(5π8)+isin(5π8))12(cos(7π8)+isin(7π8))
  3. 32(cos(π4)+isin(π4))8(cos(7π12)+isin(7π12))
  4. 4(cos(203)+isin(203))6(cos(74)+isin(74))

Solution

We will multiply and divide the complex numbers using equations ??? and ???, respectively, and then convert them to standard notation a+bi.

  1. 5(cos(11)+isin(11))8(cos(34)+isin(34))=58(cos(11+34)+isin(11+34))=40(cos(45)+isin(45))=40(22+i22)=4022+i4022=202+202i
  2. Similarly, we obtain the next product.

3(cos(5π8)+isin(5π8))12(cos(7π8)+isin(7π8))=36(cos(5π8+7π8)+isin(5π8+7π8))

Now, 5π8+7π8=5π+7π8=12π8=3π2, and cos(3π2)=0 and sin(3π2)=1. Therefore, we obtain that the product is

36(cos(3π2)+isin(3π2))=36(0+i(1))=36i

  1. For the quotient, we use the subtraction formula ???.

32(cos(π4)+isin(π4))8(cos(7π12)+isin(7π12))=328(cos(π47π12)+isin(π47π12))

Now, the difference in the argument of cos and sin is given by

π47π12=3π7π12=4π12=π3

and cos(π3)=cos(π3)=12 and sin(π3)=sin(π3)=32. With this, we obtain

32(cos(π4)+isin(π4))8(cos(7π12)+isin(7π12))=4(cos(π3)+isin(π3))=4(12i32)=223i

  1. Finally, we calculate

4(cos(203)+isin(203))6(cos(74)+isin(74))=23(cos(129)+isin(129))

Since we do not have exact values of cos and sin for the angle 129, we approximate the complex number in standard form with the calculator.

23(cos(129)+isin(129))=23cos(129)+i23sin(129)0.420+0.518i

Note that here we approximated the solution to the nearest thousandth.


This page titled 21.2: Multiplication and division of complex numbers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform.

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