How do you simplify ${\left( {2 - 2i} \right)^2}?$

Hint: The complex number consists of the real part and an imaginary part and is denoted by “Z”. It can be expressed as $z = a + ib$ where “a” is the real part and “b” is the imaginary part. An imaginary number is the complex number which can be written as the real number multiplied by the imaginary unit “i”. Here we will use the laws of the power and exponents and simplify placing the values of the squares of the imaginary number “i”.

Complete step-by-step solution:
Take the given expression: ${\left( {2 - 2i} \right)^2}$
Apply the whole square formula in the above equation using ${(a - b)^2} = {a^2} - 2ab + {b^2}$
Simplify the above expression using it.
$ = {(2)^2} - 2(2)(2i) + {(2i)^2}$
Simplify the above expression.
$ = 4 - 8i + 4{i^2}$
Now, substitute the values for “i” for ${i^2} = - 1{\text{ }}$ in the above equation
$ = 4 - 8i + 4( - 1)$
Product of one positive term and the negative term gives the resultant value negative.
$ = 4 - 8i - 4$
Make a pair of like terms in the above equation.
$ = \underline {4 - 4} - 8i$
Like terms with the same value and the opposite sign cancel each other.
$ = - 8i$
Hence the simplified value of ${\left( {2 - 2i} \right)^2}$is $( - 8i)$
This is the required solution.

Additional Information: Conjugate of the complex number is the number with an equal real part but an imaginary part equal magnitude but opposite in sign. It is expressed in the form of $a + ib$ , where “a” and “b” are the real numbers of the complex number. The modulus of the complex number is the length of the vector and can be expressed as $r = \sqrt {{a^2} + {b^2}} $.

Note: Apply the identity of the square properly. Remember while simplification of the equations always makes the pair of like terms and remember when the like terms have equal value and opposite signs then they always cancel each other.