Question

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

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

Conversion of a complex number to polar form:
Consider the complex number $x+iy$
Step 1 : Divide and multiply by $\left| x+iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
$x+iy=\sqrt{{{x}^{2}}+{{y}^{2}}}\left( \dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}+i\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right)$
Step 2: $\theta =\arctan \left( \dfrac{y}{x} \right)$
Check whether $\cos \theta =\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ and $\sin \theta =\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ or $\cos \left( \pi -\theta \right)=\dfrac{x}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$ and $\sin \left( \pi -\theta \right)=\dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$
Step 3: If $\theta$ satisfies, then $x+iy=\sqrt{{{x}^{2}}+{{y}^{2}}}\left( \cos \theta +i\sin \theta \right)$ else $x+iy=\sqrt{{{x}^{2}}+{{y}^{2}}}\left( \cos \left( \pi -\theta \right)+i\sin \left( \pi -\theta \right) \right)$
Hence the given complex number is converted in polar form.
We have $\left| 1+i \right|=\sqrt{2}$
Hence $1+i=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}} \right)$
We know that $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}=\sin \left( \dfrac{\pi }{4} \right)$
Hence, we have
Now, we have
${{\left( 1+i \right)}^{-20}}={{\left( \sqrt{2}\left( \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right) \right)}^{-20}}$
We know that ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$. Hence we have
${{\left( 1+i \right)}^{-20}}={{\left( \sqrt{2} \right)}^{-20}}{{\left( \cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right) \right)}^{-20}}$
We know that ${{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx,n\in \mathbb{Z}$
Hence we have
${{\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)}^{-20}}=\cos \left( \dfrac{\pi \left( -20 \right)}{4} \right)+i\sin \left( \dfrac{\pi }{4}\left( -20 \right) \right)=\cos \left( -5\pi \right)+i\sin \left( -5\pi \right)$
We know that $\cos \left( -x \right)=\cos x$ and $\sin \left( -x \right)=-\sin x$. Hence we have
${{\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right)}^{-20}}=\cos 5\pi -i\sin 5\pi =\cos \pi -i\sin \pi =-1$
Hence ${{\left( 1+i \right)}^{-20}}={{\left( \sqrt{2} \right)}^{-20}}\left( -1 \right)=-{{2}^{-10}}$
Hence we have
$a+ib=-{{2}^{-10}}$
Comparing real to real and imaginary to imaginary, we get
$a=-{{2}^{-10}}$ and $b=0$
Hence option [b] is correct.

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