Simplify \[{i^{19}}.\]

Hint:We know that $i = \sqrt { - 1} $.
So by using this basic equation and by finding the values of other powers of $i$we can solve this question. Here we have to convert our question in such a way that it can be expressed using various powers of $i$ whose values are known to us.

Complete step by step solution:
Given
\[{i^{19}}...............................\left( 1 \right)\]
We know that
$i = \sqrt { - 1} \;\;\; \Rightarrow {i^2} = - 1.........................\left( 2 \right)$
Now we need to find \[{i^{19}}\] which have to be expressed using various powers of $i$ whose values are known to us. So we can write:
\[{i^{19}} = {i^{\left( {16 + 3} \right)}}.................\left( 3 \right)\]
Now we know that:
${\left( x \right)^{m + n}} = {x^m} \times {x^n}...........................\left( 4 \right)$
So expressing (3) in terms of (4) we can write:
\[{i^{\left( {16 + 3} \right)}} = {i^{\left( {16} \right)}} \times {i^{\left( 3 \right)}}.......................\left( 5
\right)\]
Now we know that ${i^{\left( {16} \right)}}$can be written as:
${i^{\left( {16} \right)}} = {\left( {{i^4}} \right)^4}............\left( 6 \right)$
So we need to find the value of ${i^{\left( 4 \right)}}$and ${i^{\left( 3 \right)}}$ to simplify\[{i^{19}}\]:
Such that:
$
{i^4} = {\left( {{i^2}} \right)^2} = {\left( { - 1} \right)^2} = 1......................\left( 7 \right) \\
{i^3} = {i^2} \times i = \left( { - 1} \right) \times i = - i......................\left( 8 \right) \\
$
Now our aim is to find ${i^{19}}$, which can be expressed in terms of ${i^4}$ and ${i^3}$.
i.e. ${i^{\left( {16} \right)}} = {\left( {{i^4}} \right)^4} \times \left( {{i^3}} \right)$
So from (7) and (8) we can write the values of ${i^4}$ and$ {i^3}$.
\[
\Rightarrow {i^{19}} = {\left( 1 \right)^4} \times \left( { - i} \right) = 1 \times - i \\
\Rightarrow {i^{19}} = - i \\
\]
Therefore\[{i^{19}} = - i\].

Note:
Formulas useful for solving these types of questions:
$
i = \sqrt { - 1} \\
{i^2} = - 1\; \\
{i^3} = - i \\
{i^4} = 1 \\
{i^{4n\;}} = 1 \\
{i^{4n - 1}} = - i \\
$
So for finding higher or lower powers of $i$ we have to express its power in terms of the above powers and directly substitute the values associated with it since any power can be expressed in terms of the above equations.
Another property widely popular and used for solving imaginary powers are the exponential properties, which form the basis for solving many problems with imaginary numbers.
Complex numbers which are numbers expressed in the form of $a + bi$ where ‘a’ is the real part and ‘b’ is the imaginary part is the field where imaginary numbers are important and are of great use.