Question

The value of ${{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}$ is
$\begin{align}
  & a)8 \\
 & b)8i \\
 & c)-8 \\
 & d)32 \\
\end{align}$

Answer 1

Hint: Now first we will consider ${{\left( 1+i \right)}^{4}}$. First we will rewrite the expression with the help of property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . Now we will expand the bracket with the help of formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Now we know that ${{i}^{2}}=-1$ Hence using this we will easily get the value of ${{\left( 1+i \right)}^{4}}$ . In the same way we will expand the expression ${{\left( 1-i \right)}^{4}}$ . Now we will add the two expressions and hence find the value of ${{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}$ .

Complete step by step answer:
Now we are given a complex expression ${{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}$ .
Now any complex number z is written in the form $z=a+ib$ where a and b is the real numbers and $i=\sqrt{-1}$ .
Now $1+i$ and $1-i$ are the two complex numbers where $i=\sqrt{-1}$ .
Let us find find the values of ${{\left( 1+i \right)}^{4}}$ and ${{\left( 1-i \right)}^{4}}$
Now let us first consider ${{\left( 1+i \right)}^{4}}$
We know the property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$
Hence we can write the expression as ${{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Hence using this we get, ${{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}={{\left( 1+{{i}^{2}}+2i \right)}^{2}}$
Now we have ${{i}^{2}}=-1$ . Substituting the value in the expression we get
\[\begin{align}
  & {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}={{\left( 1-1+2i \right)}^{2}} \\
 & \Rightarrow {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}={{\left( 2i \right)}^{2}} \\
 & \Rightarrow {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}=4{{i}^{2}} \\
 & \Rightarrow {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}=-4..................................\left( 1 \right) \\
\end{align}\]
Now let us consider ${{\left( 1-i \right)}^{4}}$ .
Again using the property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$
Hence we can write the expression as ${{\left( {{\left( 1-i \right)}^{2}} \right)}^{2}}$
Now we know that ${{\left( ab \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Using this formula in the above expression we get, ${{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}={{\left( 1+{{i}^{2}}-2i \right)}^{2}}$
Now we have ${{i}^{2}}=-1$ . Let us substitute this value in the expression obtained. Hence we get,
\[\begin{align}
  & {{\left( {{\left( 1-i \right)}^{2}} \right)}^{2}}={{\left( 1-1-2i \right)}^{2}} \\
 & \Rightarrow {{\left( {{\left( 1-i \right)}^{2}} \right)}^{2}}={{\left( -2i \right)}^{2}} \\
 & \Rightarrow {{\left( {{\left( 1-i \right)}^{2}} \right)}^{2}}=4{{i}^{2}} \\
 & \Rightarrow {{\left( {{\left( 1-i \right)}^{2}} \right)}^{2}}=-4..................................\left( 2 \right) \\
\end{align}\]
Now let us add equation (1) and equation (2),
\[\begin{align}
  & {{\left( {{\left( 1+i \right)}^{2}} \right)}^{2}}+{{\left( {{\left( 1-i \right)}^{2}} \right)}^{2}}=-4+\left( -4 \right) \\
 & \Rightarrow {{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}=-8 \\
\end{align}\]
Hence we get the value of ${{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}$ is – 8.

So, the correct answer is “Option a”.

Note: Now note that despite the negative sign in the second expression we get the same values for both the expressions, which is – 4 because the square of the negative number is positive. Also we have ${{i}^{2}}=-1$ .