Question

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

Answer 1

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 $2$.