Express the complex number ${{i}^{-39}}$ in the form a + ib.

Hint: Simply write ${{\left( i \right)}^{-39}}$ as $\dfrac{1}{{{\left( i \right)}^{39}}}$ then multiply by ‘i’ to both numerator and denominator and finally we use the fact ${{\left( i \right)}^{4}}=1$

Complete Step-by-Step solution:
We have to express ${{\left( i \right)}^{39}}$ in the form of a + ib.
Before doing so, we will learn what complex numbers are.
A complex number is a number that can be written in form of a + bi, where a, b are real numbers and i is a solution of the equation ${{x}^{2}}=-1$ .This is because no real value satisfies for equation ${{x}^{2}}+1=0$ or ${{x}^{2}}=-1$ , hence i is called imaginary number. For the complex number a + ib, a is considered as real part and b as imaginary part. Despite the historical nomenclature “imaginary” complex numbers are regarded in the mathematical sciences as just as “real” as real numbers and are fundamental in any aspects of scientific description of the natural world
Now it is given that,
${{\left( i \right)}^{-39}}$
which can be written as,
$\dfrac{1}{{{\left( i \right)}^{39}}}$
Now we will multiply ‘i' to both numerator and denominator so we get,
$\dfrac{1}{{{\left( i \right)}^{40}}}$
which also be written as,
$\dfrac{1}{{{\left( {{i}^{4}} \right)}^{10}}}$
Now as we know that \[{{i}^{2}}=-1\] so, \[{{i}^{4}}=1\] now using it we get,
$\dfrac{1}{{{\left( i \right)}^{10}}}$
So, ${{\left( i \right)}^{-39}}=i=0+1i$
Hence the answer is 0 + 1i

Note: One can also it by writing ${{\left( i \right)}^{-39}}$ as ${{\left( i \right)}^{-36}}\times {{\left( i \right)}^{-3}}$ then use the fact ${{\left( i \right)}^{4}}={{\left( i \right)}^{-4}}=1$ then write ${{\left( i \right)}^{-3}}$ as ${{\left( i \right)}^{-3}}\times {{\left( i \right)}^{4}}$ because ${{i}^{4}}=1$ and multiplying by 1 does not change value to finally get answer.