Evaluate : $ {i^{-50}} $

Hint: We have to find the value of $ {i^{-50}} $ . We solve this question using the concept of complex numbers . We should also have the knowledge about the values of the powers of iota ( i ) . Firstly we write the term of iota in simplest terms and write its power in terms of \[4n{\text{ }} + {\text{ }}a\]format . And hence using the values of the power of $ i $ , we evaluate the value of $ {i^{-50}} $ .

Complete step-by-step answer:
Given :
To evaluate $ {i^{-50}} $

The expression can also be written as ,
 $ {i^{-50}} = \dfrac{1}{{[{i^{(50)}}]}} $
Now , we also know that
 $ i = \sqrt {( - 1)} $
The various values of powers of i are given as :
 $ {i^2} = - 1 $
 $ {i^3} = - i $
 $ {i^4} = 1 $
 $ {i^5} = i $
 $ {i^6} = - i $
From the values we conclude that the value of power to i repeats after in \[\left( {{\text{ }}4n{\text{ }} + {\text{ }}z{\text{ }}} \right)\] terms , where \[z{\text{ }} = {\text{ }}\left\{ {{\text{ }}1{\text{ }},{\text{ }}2{\text{ }},{\text{ }}3{\text{ }}} \right\}\]
Using these values and concept , we get
 $ {i^{-50}} = \dfrac{1}{{[{i^{(4 \times 12 + 2)}}]}} $
 $ {i^{-50}} = \dfrac{1}{{[{i^{(2)}}]}} $
As , we know $ {i^2} = - 1 $
 $ {i^{-50}} = - 1 $
Hence , we evaluate that the value of $ {i^{-50}} $ is \[ - 1\].
So, the correct answer is “-1”.

Note: A number of the form\[a{\text{ }} + {\text{ }}i{\text{ }}b\], where $ a $ and $ b $ are real numbers , is called a complex number , a is called the real part and b is called the imaginary part of the complex number .
Every real number can be represented in terms of complex numbers but every complex number can’t be represented as a real number . Also every real number is a complex number as every real number can be represented in terms of complex numbers by adding the term of iota with a constant zero multiplied to it .
Since $ {b^2} - 4ac $ determines whether the quadratic equation $ a{x^2} + bx + c = 0 $
If $ {b^2} - 4ac < 0 $ then the equation has imaginary roots .