Evaluate the expression: ${\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8}$ ?

Hint: we will simplify it as ${\left[ {{{\left\{ a \right\}}^2}} \right]^2} + {\left[ {{{\left\{ b \right\}}^2}} \right]^2}$ by substituting ${\left( {1 + i} \right)^2} = a$ and ${\left( {1 - i} \right)^2} = b$ and then we will calculate the value of a , and evaluate the value of a in terms of i to reach its power to $8$. We do the same with b and then substitute their value.

Complete step-by-step solution:
We assume of ${\left( {1 + i} \right)^2}$ as a and try to simplify,
${\left( {1 + i} \right)^2} = a$
$1 + {i^2} + 2i = a$
We substitute I square as -1,
$1 - 1 + 2i = a$
$2i = a$
We will repeat similar steps for b,
${\left( {1 - i} \right)^2} = b$
$1 + {i^2} - 2i = b$
So, $ - 2i = b$
Now, we will try to find ${\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8}$ in terms of ${a^4}$ and ${b^4}$ ,
${\left( {1 + i} \right)^8} + {\left( {1 - i} \right)^8}$
We can also write above equation as,
$ = {\left[ {{{\left\{ {{{\left( {1 + i} \right)}^2}} \right\}}^2}} \right]^2} + {\left[ {{{\left\{ {{{\left( {1 - i} \right)}^2}} \right\}}^2}} \right]^2}$
We have substituted the value a and b,
$ = {\left[ {{{\left\{ a \right\}}^2}} \right]^2} + {\left[ {{{\left\{ b \right\}}^2}} \right]^2}$
$ = {\left[ {{{\left\{ {2i} \right\}}^2}} \right]^2} + {\left[ {{{\left\{ { - 2i} \right\}}^2}} \right]^2}$
$ = {\left( {2i} \right)^4} + {\left( { - 2i} \right)^4}$
By further evaluation,
$ = 16{i^4} + 16{i^4}$
$ = 32$

Note: Students may make a mistake when determining the question's value because the powering a to power by 4 leads to our question. Complex numbers are the numbers that are expressed in the form of \[a + ib\;\] where, a, b are real numbers and ‘i’ is an imaginary number called “iota”. The value of \[i = (\sqrt 1 )\] . For example, \[2 + 3i\] is a complex number, where \[2\] is a real number (Re) and \[3i\] is an imaginary number. We should know the value of \[{i^2} = - 1\] and ${i^4} = 1$ .