How do you simplify ${{\left( 8-3i \right)}^{2}}$ ?

Hint: We know that i is an imaginary number the value of i is equal to square root of 1 so the value of ${{i}^{2}}$ is equal to - 1. We can find the square of 8 – 3i by using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and ${{i}^{2}}$ is equal to 1 .

Complete step by step answer:
We have to simplify ${{\left( 8-3i \right)}^{2}}$
We know the algebraic formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ so we can assume a equal to 8 and b equal 3i and apply the formula
So we can write ${{\left( 8-3i \right)}^{2}}={{8}^{2}}-2\times 8\times 3i+{{\left( 3i \right)}^{2}}$
The value of ${{8}^{2}}$ is 64 , and ${{\left( 3i \right)}^{2}}=-9$
Now we can write ${{\left( 8-3i \right)}^{2}}=64-48i-9$
Further solving we get
${{\left( 8-3i \right)}^{2}}=55-48i$
So the value of ${{\left( 8-3i \right)}^{2}}$ is $55-48i$

Note:
We know the value of ${{i}^{2}}$ is equal to -1. The value of ${{i}^{3}}=-i$ and the value of ${{i}^{4}}=1$ so the power of i repeated every four numbers, i to the power 5 is equal to i to power 1. So if we have to calculate ${{i}^{x}}$ then first divide x by 4, then note the remainder if the remainder is 0 then ${{i}^{x}}$ is equal to 1 , if the remainder is 2 then the value of ${{i}^{x}}$ is equal to -1, if the remainder is 3 then the value of ${{i}^{x}}$ is equal to - i . We can write any complex number in the form of $a{{e}^{i\theta }}$ , According to the Euler’s formula we can write x + iy as $a{{e}^{i\theta }}$ where $\theta $ is the polar coordinate $\theta $ of the point (x, y) and the value of a is $\sqrt{{{x}^{2}}+{{y}^{2}}}$