Question

Solve the following expression, ${{i}^{9}}+{{i}^{19}}$.

Answer 1

Hint: In order to solve this question, we should know a few properties of, $i=\sqrt{-1}$, that are, ${{i}^{4n+1}}=i;{{i}^{4n+2}}={{i}^{2}}=-1;{{i}^{4n+3}}={{i}^{3}}=-i$ and ${{i}^{4n}}=1$. Now, to solve this question, we will try to write 9 and 19 in any one of the 4n, 4n + 1, 4n + 2 or 4n + 3 forms. Then we will simplify and get the desired result.

Complete step-by-step answer:
In this question, we have been asked to solve the expression, ${{i}^{9}}+{{i}^{19}}$.
To solve this, we should know that, $i=\sqrt{-1}$, and, ${{i}^{4n+1}}=i;{{i}^{4n+2}}={{i}^{2}}=-1;{{i}^{4n+3}}={{i}^{3}}=-i$ and ${{i}^{4n}}=1$. We will solve this question by using these concepts.
So, to apply this property, we will try to express 9 and 19 in any of the 4n, 4n + 1, 4n + 2 or 4n + 3 forms.
We know that 9 = 8 + 1 = 4 (2) + 1 and 19 can be written as 16 + 3 = 4 (4) + 3. So, we will apply it in the given expression and write it as follows,
$\begin{aligned} & {{i}^{9}}+{{i}^{19}} \\\ & ={{i}^{4\left( 2 \right)+1}}+{{i}^{4\left( 4 \right)+3}} \\\ \end{aligned}$
Now, we can see that ${{i}^{9}}$ is expressed in the form of ${{i}^{4n+1}}$, which gives ${{i}^{9}}=i$.
And we can see that ${{i}^{19}}$ is expressed in the form of ${{i}^{4n+3}}$, which gives ${{i}^{19}}=-i$.
Hence, we get the given expression as, $i-i$. And we know that equal terms of opposite signs gets cancelled, so we get the expression as 0.
Hence, we can say that ${{i}^{9}}+{{i}^{19}}=0$.

Note: While solving this question, we need to remember that $i=i,{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. Sometimes we may end up with silly mistakes and write $i=-1,{{i}^{2}}=1,{{i}^{3}}=1$, which will lead to incorrect answers. We can also start by taking ${{i}^{9}}$ outside as ${{i}^{9}}\left( 1+{{i}^{10}} \right)$. Then we first compute ${{i}^{10}}$ as ${{i}^{4\left( 2 \right)+2}}=-1$, so we will get, ${{i}^{9}}\left( 1-1 \right)=0$.