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Question: Solve \({{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}=\)...

Solve (1+i)4+(1−i)4={{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}=

Explanation

Solution

We will use the general algebra of simplification to solve this question; also we use the values of iota with different powers to find the value of given expression. The following algebraic formulas will be used to simplify the given expression-
(a+b)2=a2+b2+2ab (a−b)2=a2+b2−2ab \begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ \end{aligned}

Complete step by step answer:
We have been given an expression (1+i)4+(1−i)4{{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}.
We have to find the value of the given expression.
To find the value of the given expression first let us simplify the expression we have
⇒[(1+i)2]2+[(1−i)2]2\Rightarrow {{\left[ {{\left( 1+i \right)}^{2}} \right]}^{2}}+{{\left[ {{\left( 1-i \right)}^{2}} \right]}^{2}}
Now we know that the algebraic identities of simplification are
(a+b)2=a2+b2+2ab (a−b)2=a2+b2−2ab \begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ \end{aligned}
So, when we apply the formula on the given expression we get
⇒[1+i2+2i]2+[1+i2−2i]2\Rightarrow {{\left[ 1+{{i}^{2}}+2i \right]}^{2}}+{{\left[ 1+{{i}^{2}}-2i \right]}^{2}}
Now, we know that the value of imaginary number iota with power will be i2=−1{{i}^{2}}=-1
So, by putting the value of i2{{i}^{2}} in the above equation we get
⇒[1+(−1)+2i]2+[1+(−1)−2i]2 ⇒[1−1+2i]2+[1−1−2i]2 \begin{aligned} & \Rightarrow {{\left[ 1+\left( -1 \right)+2i \right]}^{2}}+{{\left[ 1+\left( -1 \right)-2i \right]}^{2}} \\\ & \Rightarrow {{\left[ 1-1+2i \right]}^{2}}+{{\left[ 1-1-2i \right]}^{2}} \\\ \end{aligned}
Now, solving further we get
⇒[2i]2+[−2i]2 ⇒4i2+4i2 \begin{aligned} & \Rightarrow {{\left[ 2i \right]}^{2}}+{{\left[ -2i \right]}^{2}} \\\ & \Rightarrow 4{{i}^{2}}+4{{i}^{2}} \\\ \end{aligned}
Now, again putting the value i2=−1{{i}^{2}}=-1 in the above equation we get
⇒4(−1)+4(−1) ⇒−4−4 ⇒−8 \begin{aligned} & \Rightarrow 4\left( -1 \right)+4\left( -1 \right) \\\ & \Rightarrow -4-4 \\\ & \Rightarrow -8 \\\ \end{aligned}

So, we get (1+i)4+(1−i)4=−8{{\left( 1+i \right)}^{4}}+{{\left( 1-i \right)}^{4}}=-8

Note: To solve these types of questions we need to know the basics of iota ii that it is used to represent the imaginary part of a complex number of the form a+iba+ib, where a&ba\And b are real numbers and ii is the imaginary number. Also the values of different powers of ii are different so one must have knowledge about the values. The alternative way to solve this question is by applying the direct formula of a4+b4{{a}^{4}}+{{b}^{4}}. The simplified form of a4+b4{{a}^{4}}+{{b}^{4}} is as follows:
a4+b4=(a2+b2)2−2a2b2{{a}^{4}}+{{b}^{4}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}-2{{a}^{2}}{{b}^{2}}.