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Question: If \({{\left( 1+i \right)}^{-20}}=a+ib\), then the values of a and b are [a] \(a={{2}^{-10}},b=-{{...

If (1+i)−20=a+ib{{\left( 1+i \right)}^{-20}}=a+ib, then the values of a and b are
[a] a=2−10,b=−2−10a={{2}^{-10}},b=-{{2}^{-10}}
[b] a=−2−10,b=0a=-{{2}^{-10}},b=0
[c] a=2−10,b=0a={{2}^{-10}},b=0
[d] None of the above.

Explanation

Solution

Hint: convert 1+i1+i in polar form, i.e. r(cos⁡x+isin⁡x)r\left( \cos x+i\sin x \right) form and use De-Movire's formula, i.e. (cos⁡x+isin⁡x)n=cos⁡nx+isin⁡nx,n∈Z{{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx,n\in \mathbb{Z}. Alternatively, use Euler's identity, i.e. eix=cos⁡x+isin⁡x{{e}^{ix}}=\cos x+i\sin x to simplify the expression. Use the fact that cos⁡(π4)=sin⁡(π4)=12\cos \left( \dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}.
Hence find the value of the above expression.

Complete step-by-step answer:
Conversion of a complex number to polar form:
Consider the complex number x+iyx+iy
Step 1 : Divide and multiply by ∣x+iy∣=x2+y2\left| x+iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}
x+iy=x2+y2(xx2+y2+iyx2+y2)x+iy=\sqrt{{{x}^{2}}+{{y}^{2}}}\left( \dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}+i\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right)
Step 2: θ=arctan⁡(yx)\theta =\arctan \left( \dfrac{y}{x} \right)
Check whether cos⁡θ=xx2+y2\cos \theta =\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}} and sin⁡θ=yx2+y2\sin \theta =\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}} or cos⁡(π−θ)=xx2+y2\cos \left( \pi -\theta \right)=\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}} and sin⁡(π−θ)=yx2+y2\sin \left( \pi -\theta \right)=\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}
Step 3: If θ\theta satisfies, then x+iy=x2+y2(cos⁡θ+isin⁡θ)x+iy=\sqrt{{{x}^{2}}+{{y}^{2}}}\left( \cos \theta +i\sin \theta \right) else x+iy=x2+y2(cos⁡(π−θ)+isin⁡(π−θ))x+iy=\sqrt{{{x}^{2}}+{{y}^{2}}}\left( \cos \left( \pi -\theta \right)+i\sin \left( \pi -\theta \right) \right)
Hence the given complex number is converted in polar form.
We have ∣1+i∣=2\left| 1+i \right|=\sqrt{2}
Hence 1+i=2(12+i2)1+i=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right)
We know that cos⁡(π4)=12=sin⁡(π4)\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}=\sin \left( \dfrac{\pi }{4} \right)
Hence, we have
Now, we have
(1+i)−20=(2(cos⁡(π4)+isin⁡(π4)))−20{{\left( 1+i \right)}^{-20}}={{\left( \sqrt{2}\left( \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right) \right)}^{-20}}
We know that (ab)n=anbn{{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}. Hence we have
(1+i)−20=(2)−20(cos⁡(π4)+isin⁡(π4))−20{{\left( 1+i \right)}^{-20}}={{\left( \sqrt{2} \right)}^{-20}}{{\left( \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right)}^{-20}}
We know that (cos⁡x+isin⁡x)n=cos⁡nx+isin⁡nx,n∈Z{{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx,n\in \mathbb{Z}
Hence we have
(cos⁡π4+isin⁡π4)−20=cos⁡(π(−20)4)+isin⁡(π4(−20))=cos⁡(−5π)+isin⁡(−5π){{\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)}^{-20}}=\cos \left( \dfrac{\pi \left( -20 \right)}{4} \right)+i\sin \left( \dfrac{\pi }{4}\left( -20 \right) \right)=\cos \left( -5\pi \right)+i\sin \left( -5\pi \right)
We know that cos⁡(−x)=cos⁡x\cos \left( -x \right)=\cos x and sin⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x. Hence we have
(cos⁡π4+isin⁡π4)−20=cos⁡5π−isin⁡5π=cos⁡π−isin⁡π=−1{{\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)}^{-20}}=\cos 5\pi -i\sin 5\pi =\cos \pi -i\sin \pi =-1
Hence (1+i)−20=(2)−20(−1)=−2−10{{\left( 1+i \right)}^{-20}}={{\left( \sqrt{2} \right)}^{-20}}\left( -1 \right)=-{{2}^{-10}}
Hence we have
a+ib=−2−10a+ib=-{{2}^{-10}}
Comparing real to real and imaginary to imaginary, we get
a=−2−10a=-{{2}^{-10}} and b=0b=0
Hence option [b] is correct.

Note: [1] Alternative solution:
1+i=2(cos⁡(π4)+isin⁡(π4))1+i=\sqrt{2}\left( \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right)
We know that cos⁡x+isin⁡x=eix\cos x+i\sin x={{e}^{ix}}
Put x=π4x=\dfrac{\pi }{4} in the above formula, we get
cos⁡π4+isin⁡π4=eiπ4\cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}={{e}^{\dfrac{i\pi }{4}}}
Hence we have
1+i=2eiπ41+i=\sqrt{2}{{e}^{\dfrac{i\pi }{4}}}
Raising power to -20 on both sides, we get
(1+i)−20=(2eiπ4)−20=(2)−20e−i5π{{\left( 1+i \right)}^{-20}}={{\left( \sqrt{2}{{e}^{\dfrac{i\pi }{4}}} \right)}^{-20}}={{\left( \sqrt{2} \right)}^{-20}}{{e}^{-i5\pi }}
We know that ei4π=1{{e}^{i4\pi }}=1 and eiπ=−1{{e}^{i\pi }}=-1.
Hence we have
(1+i)−20=−2−10{{\left( 1+i \right)}^{-20}}=-{{2}^{-10}}, which is the same as obtained above.
Hence option [b] is correct.
[2] Alternative solution:
We know that ∣zn∣=∣z∣n\left| {{z}^{n}} \right|={{\left| z \right|}^{n}} and arg⁡(zn)=narg⁡(z)\arg \left( {{z}^{n}} \right)=n\arg \left( z \right)
Hence we have ∣(1+i)−20∣=(2)−20=2−10\left| {{\left( 1+i \right)}^{-20}} \right|={{\left( \sqrt{2} \right)}^{-20}}={{2}^{-10}} and arg⁡((1+i)−20)=−20arg⁡(1+i)=−20π4=−5π\arg \left( {{\left( 1+i \right)}^{-20}} \right)=-20\arg \left( 1+i \right)=-20\dfrac{\pi }{4}=-5\pi
We know that z=∣z∣(cos⁡(arg⁡z)+isin⁡(arg⁡z))z=\left| z \right|\left( \cos \left( \arg z \right)+i\sin \left( \arg z \right) \right)
Hence we have
(1+i)−20=2−10(cos⁡(−5π)+isin⁡(−5π))=−2−10{{\left( 1+i \right)}^{-20}}={{2}^{-10}}\left( \cos \left( -5\pi \right)+i\sin \left( -5\pi \right) \right)=-{{2}^{-10}}, which is the same as obtained above.
Hence option [b] is correct.