Respuesta :
[tex]56(cos(33^{\circ}) + i sin(33^{\circ})) \div 7(cos(11^{\circ}) + i sin(11^{\circ}))\\\\=8(cos(22^{\circ}) + i~ sin(22^{\circ}))[/tex]
What is complex number?
"The number of the form a + ib, where a, b are real numbers and [tex]i=\sqrt{-1}[/tex]"
What is De Moivre's theorem?
"This theorem gives a formula for computing powers of complex numbers.
[tex](r(cos\theta+i~sin\theta))^n=r^n~(cos(n\theta)+i~sin(n\theta))[/tex] "
For given question,
We need to divide two complex numbers.
[tex]56(cos(33^{\circ}) + i sin(33^{\circ})) \div 7(cos(11^{\circ}) + i sin(11^{\circ}))[/tex]
Consider,
[tex]56(cos(33^{\circ}) + i~ sin(33^{\circ})) \div 7(cos(11^{\circ}) + i ~sin(11^{\circ}))\\\\=8[(cos(33^{\circ}) + i~ sin(33^{\circ})) \div (cos(11^{\circ}) + i ~sin(11^{\circ}))][/tex] ...........(i)
Consider the first complex number.
[tex](cos(33^{\circ}) + i sin(33^{\circ}))[/tex]
We can write this complex number as,
[tex]cos(3(11^{\circ}) )+ i sin(3(11^{\circ}))[/tex]
By De Moivre's theorem,
[tex]cos(3(11^{\circ}) )+ i~ sin(3(11^{\circ}))\\\\=(cos(11^{\circ})+ i~ sin(11^{\circ}))^3[/tex]
Substitute this value in (i),
[tex]56(cos(33^{\circ}) + i~ sin(33^{\circ})) \div 7(cos(11^{\circ}) + i ~sin(11^{\circ}))\\\\=8[(cos(33^{\circ}) + i~ sin(33^{\circ})) \div (cos(11^{\circ}) + i ~sin(11^{\circ}))]\\\\=8[(cos(11^{\circ}) + i~ sin(11^{\circ}))^3 \div 7(cos(11^{\circ}) + i~ sin(11^{\circ}))]\\\\=8[(cos(11^{\circ}) + i~ sin(11^{\circ}))^2]\\\\=8[(cos(2\times 11^{\circ}) + i~ sin(2\times 11^{\circ}))]~~~~~~~~~.............(De~Moivre's~theorem )\\\\=8(cos(22^{\circ}) + i~ sin(22^{\circ}))[/tex]
Therefore, [tex]56(cos(33^{\circ}) + i sin(33^{\circ})) \div 7(cos(11^{\circ}) + i sin(11^{\circ}))[/tex] is [tex]8(cos(22^{\circ}) + i~ sin(22^{\circ}))[/tex]
Learn more about the De Moivre's theorem here:
brainly.com/question/17211848
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