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Question: Find \({(1 + i)^5}\), using DeMoivre’s theorem....

Find (1+i)5{(1 + i)^5}, using DeMoivre’s theorem.

Explanation

Solution

This problem is from complex numbers and we need to know the DeMoivre’s theorem to solve this equation. First we have to consider the given equation and we have to equate it to a general form and then we have to consider the power and by De Moivre’s theorem, we need to solve this problem.

Complete answer:
For any rational number, (cos⁡nθ+isin⁡nθ)(\cos n\theta + i\sin n\theta ) is the value or one of the values for (cos⁡θ+isin⁡θ)n{(\cos \theta + i\sin \theta )^n}. Write the given equation in polar form i.e.., as r(cos⁡θ+isin⁡θ)r(\cos \theta + i\sin \theta ) and equate the real and imaginary parts. First let’s consider the question without considering the power value which is,
(1+i)=r(cos⁡θ+isin⁡θ) (1+i)=rcos⁡θ+rsin⁡θ)  (1 + i) = r(\cos \theta + i\sin \theta ) \\\ (1 + i) = r\cos \theta + r\sin \theta ) \\\
Equating real and imaginary term, we get,
rcos⁡θ=1r\cos \theta = 1 ………………………………………………….. (1)
rsin⁡θ=1r\sin \theta = 1 ………………………………………………..… (2)
Squaring and adding the (1) and (2) equation we get,
  ⟹  (r2cos⁡2θ+r2sin⁡2θ)=1+1   ⟹  (r2cos⁡2θ+r2sin⁡2θ)=2   ⟹  r2(cos⁡2θ+sin⁡2θ)=2   ⟹  r2=2 r=2  \implies ({r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta ) = 1 + 1 \\\ \implies ({r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta ) = 2 \\\ \implies {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 2 \\\ \implies {r^2} = 2 \\\ r = \sqrt 2 \\\
Substituting the value of r in equation (1) and (2) we get,
cos⁡θ=12 sin⁡θ=12  \cos \theta = \dfrac{1}{{\sqrt 2 }} \\\ \sin \theta = \dfrac{1}{{\sqrt 2 }} \\\
And the value of θ=π4\theta = \dfrac{\pi }{4}, for both, now let’s substitute the value of rr and θ\theta in the equation
(1+i)5=(2cos⁡θ+2isin⁡θ)5{(1 + i)^5} = {(\sqrt 2 \cos \theta + \sqrt 2 i\sin \theta )^5}, we get,
(1+i)5=(2)5(cos⁡π4+isin⁡π4)5{(1 + i)^5} = {(\sqrt 2 )^5}{(\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4})^5}
According to DeMoivre’s theorem, (cos⁡θ+isin⁡θ)n{(\cos \theta + i\sin \theta )^n} == (cos⁡nθ+isin⁡nθ)(\cos n\theta + i\sin n\theta )
(1+i)5=(2)5(cos⁡5π4+isin⁡5π4)5{(1 + i)^5} = {(\sqrt 2 )^5}{(\cos \dfrac{{5\pi }}{4} + i\sin \dfrac{{5\pi }}{4})^5}
(1+i)5=(2)5[cos⁡(π+π4)+i(sin⁡(π+π4)] (1+i)5=(2)5[−cos⁡π4−isin⁡(π4)] (1+i)5=(2)5(−12−i12) (1+i)5=−(2)4−i(2)4 (1+i)5=−4−i4  {(1 + i)^5} = {(\sqrt 2 )^5}[\cos (\pi + \dfrac{\pi }{4}) + i(\sin (\pi + \dfrac{\pi }{4})] \\\ {(1 + i)^5} = {(\sqrt 2 )^5}[ - \cos \dfrac{\pi }{4} - i\sin (\dfrac{\pi }{4})] \\\ {(1 + i)^5} = {(\sqrt 2 )^5}( - \dfrac{1}{{\sqrt 2 }} - i\dfrac{1}{{\sqrt 2 }}) \\\ {(1 + i)^5} = - {(\sqrt 2 )^4} - i{(\sqrt 2 )^4} \\\ {(1 + i)^5} = - 4 - i4 \\\
This is the required solution.

Additional Information: (cos⁡θ+isin⁡θ)n{(\cos \theta + i\sin \theta )^n} == (cos⁡nθ+isin⁡nθ)(\cos n\theta + i\sin n\theta ), where i is an imaginary term. This formula connects trigonometry and complex numbers. And it is considered as one of the important problems in Mathematics. This formula is not valid for non-integer powers m. However, the general formula of this formula is also used for other exponents also. This formula is valid for any real number and integer.

Note: Angle 5π2\dfrac{{5\pi }}{2} les in the third quadrant, cosine function is negative in the third quadrant, which is,
cos⁡(180+θ)=−cos⁡θ sin⁡(180+θ)=−sin⁡θ \cos (180 + \theta ) = - \cos \theta\\\ \sin (180 + \theta ) = - \sin \theta
As I already mentioned, this formula is valid for any real number and integer, the general form of this formula is also valid for non-integer.