Question

What is the value of ${{i}^{1000}}+{{i}^{1001}}+{{i}^{1002}}+{{i}^{1003}}$, where $i=\sqrt{-1}$?
(a) 0
(b) $i$
(c) $-i$
(d) 1

Answer 1

Hint: To find the value of the given expression, calculate the square, cube, and fourth power of $i=\sqrt{-1}$. Use the fact that the higher powers of $i=\sqrt{-1}$ can be written in terms of these lower powers of $i$. Simplify the given equation and rewrite the terms into simpler terms to get the final answer.

Complete step-by-step answer:
We have to calculate the value of ${{i}^{1000}}+{{i}^{1001}}+{{i}^{1002}}+{{i}^{1003}}$.
We know that $i$ is a square root of unity. Thus, we have $i=\sqrt{-1}$.
We will calculate the square, cube, and fourth power of $i=\sqrt{-1}$.
Thus, squaring the above equation, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$. Taking the cube of the equation $i=\sqrt{-1}$, we have ${{i}^{3}}={{\left( \sqrt{-1} \right)}^{3}}={{i}^{2+1}}=-1\times i=-i$. Taking the fourth power of $i=\sqrt{-1}$, we have ${{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}={{\left( -1 \right)}^{2}}=1$ .
We will now write the higher powers of $i=\sqrt{-1}$ in terms of the above calculated powers of $i=\sqrt{-1}$.
We can rewrite ${{i}^{1000}}$ as ${{i}^{1000}}={{\left( {{i}^{4}} \right)}^{250}}$. Thus, we have ${{i}^{1000}}={{\left( {{i}^{4}} \right)}^{250}}={{\left( 1 \right)}^{250}}=1$.
Similarly, we will write ${{i}^{1001}}$ in terms of lower powers of $i$ . Thus, we have ${{i}^{1001}}={{i}^{4\times 250+1}}={{\left( {{i}^{4}} \right)}^{250}}\times i=1\times i=i$ .
We will now rewrite ${{i}^{1002}}$ in terms of lower powers of $i$. Thus, we have ${{i}^{1002}}={{i}^{4\times 250+2}}={{\left( 1 \right)}^{250}}\times {{i}^{2}}=1\times \left( -1 \right)=-1$.
We will rewrite ${{i}^{1003}}$ in terms of lower powers of $i$. Thus, we have ${{i}^{1003}}={{i}^{4\times 250+3}}={{\left( {{i}^{4}} \right)}^{250}}\times {{i}^{3}}=1\times \left( -i \right)=-i$ .
Thus, we can rewrite ${{i}^{1000}}+{{i}^{1001}}+{{i}^{1002}}+{{i}^{1003}}$ as ${{i}^{1000}}+{{i}^{1001}}+{{i}^{1002}}+{{i}^{1003}}=1+i+\left( -1 \right)+\left( -i \right)=0$.
Hence, the value of the expression ${{i}^{1000}}+{{i}^{1001}}+{{i}^{1002}}+{{i}^{1003}}$ is 0, which is option (a).

Note: It’s necessary to write the higher powers of $i$ in terms of lower powers to simplify the given expression. Otherwise, we won’t be able to solve this question. $i$ represents the square root of unity. It is the root of the equation ${{x}^{2}}+1=0$. It denotes the imaginary part of complex numbers.