Solveeit Logo

Question

Question: How do you simplify \({{i}^{17}}\)?...

How do you simplify i17{{i}^{17}}?

Explanation

Solution

We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities and then we use the imaginary value and find all the conditions related to ii. We find the simplified form of i17{{i}^{17}}.

Complete step-by-step solution:
We know the exponent form of the number aa with the exponent being nn can be expressed as an{{a}^{n}}. In case the value of nn becomes negative, the value of the exponent takes its inverse value. The formula to express the form is an=1an,nR+{{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}.
If we take two exponential expressions where the exponents are mm and nn.
Let the numbers be am{{a}^{m}} and an{{a}^{n}}. We take multiplication of these numbers.
The indices get added. So, am+n=am×an{{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}.
The division works in an almost similar way. The indices get subtracted. So, aman=amn\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}.
For given equation i17{{i}^{17}}, we convert it as i17=i16+1=i16×i{{i}^{17}}={{i}^{16+1}}={{i}^{16}}\times i.
We also have the identity of amn=(am)n{{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}. Therefore, i16=(i4)4{{i}^{16}}={{\left( {{i}^{4}} \right)}^{4}}.
We have the relations for imaginary ii where i2=1,i3=i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1. We place the values and get
i16=(i4)4=14=1{{i}^{16}}={{\left( {{i}^{4}} \right)}^{4}}={{1}^{4}}=1. This gives i17=1×i=i{{i}^{17}}=1\times i=i
Therefore, the simplified form of i17{{i}^{17}} is ii.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices. For numbers am{{a}^{m}} and an{{a}^{n}}, we have am±an=am(1±anm){{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right).the relation is independent of the values of mm and nn. We need to remember that the condition for am=anm=n{{a}^{m}}={{a}^{n}}\Rightarrow m=n is that the value of a0,±1a\ne 0,\pm 1.