How do you write the expression ${i^{ - 18}}$ in the standard form $a + bi?$

Hint: As we know that a complex number is a number which can be expressed in the $a + bi$ form, where $a$ and $b$ are real numbers and $i$ is the imaginary number. It means it consists of both real and imaginary part. Now we have to convert the given number which is also in exponential form to the form of complex number i.e. $a + bi$.

Complete step by step answer:
As per the given question we have an exponential number ${i^{ - 18}}$.
First we have to convert the exponential form in the normal number. We know that the value of $i$ is $\sqrt { - 1} $.
So ${i^{ - 1}}$ can be written as $\dfrac{1}{{\sqrt { - 1} }}$. Now we square it: ${i^{ - 2}} = {\left( {\dfrac{1}{{\sqrt { - 1} }}} \right)^2}$. It gives us the value ${i^{ - 2}} = \dfrac{1}{{ - 1}} = - 1$.
We can write ${i^{ - 18}}$ as the form of ${({i^{ - 2}})^9}$. Now by substituting the value we have ${( - 1)^9} = - 1$.
There is no such value of $b$ given, so we assume it is zero. Now the exponential part can be written as $( - 1) + (0)i$.
Hence the standard form of ${i^{ - 18}}$ is $( - 1) + (0)i$.

Note: We should be careful while calculating the values and in the square of the imaginary part we should note that when the power is negative then the value inside remains negative. The imaginary part of the complex number is known as iota. We should be careful that ${i^2} = - 1$, not the positive value because the value of iota is at the root under.