Question

Find the value of ${(1+i)}^8$.

Answer 1

Hint: We use algebraic equation to simplify the question by placing \[{{\left( 1+i \right)}^{2}}=x\] and then we find the value of \[{{x}^{2}}\] and equate the value of \[{{x}^{2}}\] in terms of \[i\] with the value of \[x\] in terms of \[i\] till we achieve the power of \[8\] on both sides of L.H.S and R.H.S.

Complete step-by-step answer:
Placing the value of \[{{\left( 1+i \right)}^{2}}=x\]
\[{{\left( 1+i \right)}^{2}}=x\]
\[1+{{i}^{2}}+2i=x\]
\[1+\left( -1 \right)+2i=x\]
\[2i=x\]
Now placing the value of \[{{\left( 1+i \right)}^{2}}=x\] and squaring till \[{{\left( 1+i \right)}^{8}}\] we get:
\[{{\left( 1+i \right)}^{2}}=x\]
\[{{\left( {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}} \right)}^{2}}={{\left( {{x}^{2}} \right)}^{2}}\]
Placing the value of \[x=2i\]
\[{{\left( {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}} \right)}^{2}}={{\left( {{x}^{2}} \right)}^{2}}\]
\[{{\left( {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}} \right)}^{2}}={{\left( {{\left( 2i \right)}^{2}} \right)}^{2}}\]
\[{{\left( 1+i \right)}^{8}}={{\left( 2i \right)}^{4}}\]
\[{{\left( 1+i \right)}^{8}}=2\times 2\times 2\times 2\times i\times i\times i\times i\]
\[{{\left( 1+i \right)}^{8}}=16{{i}^{4}}\]
${{1+i}^8} = 16$ as $i^2 = -1$
Hence, the value of \[{{\left( 1+i \right)}^{8}}=16\].

Note: Students may go wrong while finding the value of the question as the L.H.S is powered by \[2\] three times while RHS is powered by \[2\] two times as the value of \[{{\left( 1+i \right)}^{2}}=x\] hence, for L.H.S base \[\left( 1+i \right)\] there are three power \[2\] and for \[x\] the power is raised by two \[2\].