Question

Evaluate the following ${{i}^{403}}$

Answer 1

Hint: Now consider the given number ${{i}^{403}}$ we know that the value of ${{i}^{4}}=1$ and ${{i}^{3}}=-i$ hence we will first simplify the power and write the number in the form of ${{i}^{3}}$ and ${{i}^{4}}$ by using the laws of indices ${{x}^{m+n}}={{x}^{m}}{{x}^{n}}$ and ${{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}}$ . Now we will substitute the values of ${{i}^{3}}$ and ${{i}^{4}}$ and hence find the solution.

Complete step by step answer:
Now let us first understand the meaning of letter i.
Now we know the number line which represents real numbers which are either rational or irrational.
But there are also numbers which are not real. These numbers are called complex numbers.
Complex numbers are numbers of the form a + ib. where a and b are real and the letter i denotes iota which is nothing but $\sqrt{-1}$ .
Now since we have $i=\sqrt{-1}$ squaring both the sides we get ${{i}^{2}}=-1$
Now multiplying I on both sides we get, ${{i}^{3}}=-i$ again multiplying i on both sides we get, ${{i}^{4}}=-i\times i=-\left( -1 \right)=1$
Hence we can say that ${{i}^{4}}=1$
Now consider the given number ${{i}^{403}}$ .
Now we know by law of indices that ${{x}^{m+n}}={{x}^{m}}{{x}^{n}}$
Hence we can write
$\Rightarrow {{i}^{403}}={{i}^{400+3}}={{i}^{400}}{{i}^{3}}$
Now again by law of indices we know that ${{x}^{mn}}={{\left( {{x}^{m}} \right)}^{n}}$ hence using this we get,
$\Rightarrow {{i}^{403}}={{\left( {{i}^{4}} \right)}^{100}}{{i}^{3}}$
Now since ${{i}^{4}}=1$ and ${{i}^{3}}=-1$ we will substitute the values in the equation,
$\Rightarrow {{i}^{403}}={{1}^{100}}\left( -i \right)=-i$
Hence the value of ${{i}^{403}}$ is – i.

Note: Note that any power to i will be either of $-1,-i,1,i$ as after ${{i}^{4}}$ same values of I will keep repeating. Hence we can find any power of i by just using the laws of indices and the known values of i. Also note that here we also get 1 and -1 as solutions which means the square of a complex number is real. Hence we can say that multiplication of two complex numbers can be real numbers.