Question

Find the value of ${{\left( 1+i \right)}^{10}}$ , where ${{i}^{2}}=-1$ , is equal to:
a). 32 i
b). 64+i
c). 24i-32
d). None of these

Answer 1

Hint: In this problem, we will use a formula in complex numbers which converts complex numbers of type $a+bi$ to the form $r{{e}^{i\theta }}$ .once we do this we will apply the law of indices which says ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and we will get the final answer.

Complete step-by-step solution -
Now the formula we are going to use is, $a+bi=\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right){{e}^{i\theta }}$ , where,
(i) $\theta ={{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in I$ quadrant.
(ii) $\theta =\pi -{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in II$ quadrant.
(iii) $\theta =\pi +{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in III$ quadrant.
(iv) $\theta =-{{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ , if $\left( a,b \right)\in IV$ quadrant.
Given a complex number is 1+i.
Comparing with $a+bi$, we get a=1 and b=1
$\therefore \left( a,b \right)\text{ that is }\left( 1,1 \right)\text{ belongs to I quadrant}$ .
Therefore, the formula is
$a+bi=\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right){{e}^{i\theta }}$ , where $\theta ={{\tan }^{-1}}\left( \left| \dfrac{b}{a} \right| \right)$ .
Thus applying the formula, we get,
$1+i=\left( \sqrt{{{1}^{2}}+{{1}^{2}}} \right){{e}^{i\theta }}$ , where $\theta ={{\tan }^{-1}}\left( \left| \dfrac{1}{1} \right| \right)$ .
$\Rightarrow 1+i=\left( \sqrt{1+1} \right){{e}^{i\theta }}$ , where $\theta ={{\tan }^{-1}}\left( 1 \right)$ .
$\Rightarrow 1+i=\sqrt{2}{{e}^{\dfrac{i\pi }{4}}}$ , Since $\theta =\dfrac{\pi }{4}$ ……………….. (i)
Now, we need to calculate the value of ${{\left( 1+i \right)}^{10}}$ .
Thus, from (i) we know that $1+i=\sqrt{2}{{e}^{\dfrac{i\pi }{4}}}$ .
Therefore,
$\begin{aligned} & {{\left( 1+i \right)}^{10}}={{\left( \sqrt{2}{{e}^{\dfrac{i\pi }{4}}} \right)}^{10}} \\\ & \Rightarrow {{\left( 1+i \right)}^{10}}={{\left( \sqrt{2} \right)}^{10}}{{\left( {{e}^{\dfrac{i\pi }{4}}} \right)}^{10}} \\\ & \Rightarrow {{\left( 1+i \right)}^{10}}={{\left( {{2}^{\dfrac{1}{2}}} \right)}^{10}}{{\left( {{e}^{\dfrac{i\pi }{4}}} \right)}^{10}} \\\ \end{aligned}$
Applying ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ , we get
${{\left( 1+i \right)}^{10}}=\left( {{2}^{\dfrac{1}{2}\times 10}} \right)\left( {{e}^{\dfrac{i\pi }{4}\times 10}} \right)$
Which is
${{\left( 1+i \right)}^{10}}=\left( {{2}^{5}} \right)\left( {{e}^{\dfrac{5\pi i}{2}}} \right)$
$\Rightarrow {{\left( 1+i \right)}^{10}}=\left( 32 \right)\left( {{e}^{\dfrac{5\pi i}{2}}} \right)$ ………….. (ii)
Now we know the Euler’s formula,
${{e}^{i\theta }}=\cos \theta +i\sin \theta$ .
Putting$\theta =\left( \dfrac{5\pi }{2} \right)$ , we get,
${{e}^{i\left( \dfrac{5\pi }{2} \right)}}=\cos \left( \dfrac{5\pi }{2} \right)+i\sin \left( \dfrac{5\pi }{2} \right)$
${{e}^{i\left( \dfrac{5\pi }{2} \right)}}=0+i\left( 1 \right)$ ……. Since odd multiples of $\left( \dfrac{\pi }{2} \right)$ give 0 in cos functions and $\sin \left( \dfrac{5\pi }{2} \right)=\sin \left( \dfrac{\pi }{2} \right)$ which is 1.
Therefore, ${{e}^{\dfrac{5\pi i}{2}}}=i$ ……………….. (iii)
Substituting (iii) in (ii) we get
${{\left( 1+i \right)}^{10}}=\left( 32 \right)\left( i \right)$
$\Rightarrow {{\left( 1+i \right)}^{10}}=32i$
Thus option (a) is correct.

Note: This problem can also be solved using binomial theorem which is
${{\left( a+b \right)}^{n}}=n{{C}_{0}}{{a}^{n}}{{b}^{0}}+n{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+n{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+............+n{{C}_{0}}{{a}^{0}}{{b}^{n}}$
Putting a=1, b=I and n=10, we get,
${{\left( 1+i \right)}^{10}}=10{{C}_{0}}{{\left( 1 \right)}^{10}}{{\left( i \right)}^{0}}+10{{C}_{1}}{{\left( 1 \right)}^{10-1}}{{\left( i \right)}^{1}}+10{{C}_{2}}{{\left( 1 \right)}^{10-2}}{{\left( i \right)}^{2}}+............+10{{C}_{10}}{{\left( 1 \right)}^{0}}{{\left( i \right)}^{10}}$
$\begin{aligned} & \Rightarrow {{\left( 1+i \right)}^{10}}=\left( 1 \right)\left( 1 \right)\left( 1 \right)+\left( 10 \right)\left( 1 \right)\left( i \right)+\left( 45 \right)\left( 1 \right)\left( {{i}^{2}} \right)+........+\left( 1 \right)\left( 1 \right)\left( {{i}^{10}} \right) \\\ & \Rightarrow {{\left( 1+i \right)}^{10}}=1+10i+45{{i}^{2}}+120{{i}^{3}}+210{{i}^{4}}+252{{i}^{5}}+210{{i}^{6}}+120{{i}^{7}}+45{{i}^{8}}+10{{i}^{9}}+{{i}^{10}} \\\ \end{aligned}$
Now given ${{i}^{2}}=-1$
Therefore, ${{i}^{3}}=-i,{{i}^{4}}=1,{{i}^{5}}=i,{{i}^{6}}=-1$ and so on.
Therefore, $\begin{aligned} & {{\left( 1+i \right)}^{10}}=1+10i-45+120\left( -i \right)+210+252i-210-120i+45+10i-1 \\\ & \Rightarrow {{\left( 1+i \right)}^{10}}=0+32i \\\ & \Rightarrow {{\left( 1+i \right)}^{10}}=32i \\\ \end{aligned}$