21.2: Multiplication and division of complex numbers
( \newcommand{\kernel}{\mathrm{null}\,}\)
One interesting feature of the polar form of a complex number is that the multiplication and division are very easy to perform.
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(θ1−θ2)+isin(θ1−θ2))
- Proof
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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(θ2−θ2)+isin(θ2−θ2))=1⋅(cos0+isin0)=1⋅(1+i⋅0)=1⟹1r2(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(θ1−θ2)+isin(θ1−θ2))
Multiply or divide the complex numbers, and write your answer in polar and standard form.
- 5(cos(11∘)+isin(11∘))⋅8(cos(34∘)+isin(34∘))
- 3(cos(5π8)+isin(5π8))⋅12(cos(7π8)+isin(7π8))
- 32(cos(π4)+isin(π4))8(cos(7π12)+isin(7π12))
- 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.
- 5(cos(11∘)+isin(11∘))⋅8(cos(34∘)+isin(34∘))=5⋅8⋅(cos(11∘+34∘)+isin(11∘+34∘))=40(cos(45∘)+isin(45∘))=40(√22+i√22)=40√22+i⋅40√22=20√2+20√2i
- 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
- For the quotient, we use the subtraction formula ???.
32(cos(π4)+isin(π4))8(cos(7π12)+isin(7π12))=328(cos(π4−7π12)+isin(π4−7π12))
Now, the difference in the argument of cos and sin is given by
π4−7π12=3π−7π12=−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⋅(12−i√32)=2−2√3⋅i
- 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∘))=23⋅cos(129∘)+i⋅23⋅sin(129∘)≈−0.420+0.518⋅i
Note that here we approximated the solution to the nearest thousandth.