Question

Find the value of
$\dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1$

Answer 1

Hint:
To solve this question first divide the powers by 4 and then split all the powers of the terms in the form of multiplication with 4. E.g. ${i^{592}} = {i^{148 \times 4}}$ and if the power is not divisible by 4 then split the power in addition of remainder when divided by 4 and the quotient with multiplication of 4. E.g. ${i^{590}} = {i^{147 \times 4 + 2}} = {i^{147 \times 4}}.{i^2}$. Now, put the value of ${i^4} = 1$, ${i^2} = - 1$in the expression and solve the expression to get the answer.

Complete step by step solution:
First of all, let's see whose value we have to find?
We have to find the value of $\dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1$ ……..(1)
To find the value of this, we have to simplify each term in numerator and denominator.
To simplify each term, we will first split the power of the terms by dividing their power with 4 and write it as multiple 4 plus remainder e.g. ${i^{592}} = {i^{148 \times 4}}$, here when we divide 592 with 4, we get 148 and the remainder is 0. So, 592 can be replaced with $148 \times 4$. Similarly, if we divide 590 by 4, we get 147 and the remainder gets 2. Hence, 590 can be replaced by $147 \times 4 + 2$ and we write the value of ${i^{590}} = {i^{147 \times 4 + 2}} = {i^{147 \times 4}}.{i^2}$. Similarly, we can write ${i^{588}} = {i^{147 \times 4}}$ , ${i^{586}} = {i^{146 \times 4 + 2}} = {i^{146 \times 4}}.{i^2}$ , ${i^{584}} = {i^{146 \times 4}}$, ${i^{582}} = {i^{145 \times 4 + 2}} = {i^{145 \times 4}}.{i^2}$, ${i^{580}} = {i^{145 \times 4}}$, ${i^{578}} = {i^{144 \times 4 + 2}} = {i^{144 \times 4}}.{i^2}$, ${i^{576}} = {i^{144 \times 4}}$, ${i^{574}} = {i^{143 \times 4 + 2}} = {i^{143 \times 4}}.{i^2}$.
Putting all these value in (1), we get,
\[ \Rightarrow \dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = \dfrac{{{i^{148 \times 4}} + {i^{147 \times 4}}.{i^2} + {i^{147 \times 4}} + {i^{146 \times 4}}.{i^2} + {i^{146 \times 4}}}}{{{i^{145 \times 4}}.{i^2} + {i^{145 \times 4}} + {i^{144 \times 4}}.{i^2} + {i^{144 \times 4}} + {i^{143 \times 4}}.{i^2}}} - 1\]
………..(2)
Now, write each term in double power as given:
$ \Rightarrow {x^{a \times b}} = {\left( {{x^a}} \right)^b} = {\left( {{x^b}} \right)^a}$
Apply this concept on all terms in equation (2), we get,
\[ \Rightarrow \dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = \dfrac{{{{\left( {{i^4}} \right)}^{148}} + {{\left( {{i^4}} \right)}^{147}}.{i^2} + {{\left( {{i^4}} \right)}^{147}} + {{\left( {{i^4}} \right)}^{146}}.{i^2} + {{\left( {{i^4}} \right)}^{146}}}}{{{{\left( {{i^4}} \right)}^{145}}.{i^2} + {{\left( {{i^4}} \right)}^{145}} + {{\left( {{i^4}} \right)}^{144}}.{i^2} + {{\left( {{i^4}} \right)}^{144}} + {{\left( {{i^4}} \right)}^{143}}.{i^2}}} - 1\] ………….…(3)
We know that of ${i^4} = 1$ and ${i^2} = - 1$, hence put these values in (3), we get,
\[ \Rightarrow \dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = \dfrac{{{{\left( 1 \right)}^{148}} + {{\left( 1 \right)}^{147}}.\left( { - 1} \right) + {{\left( 1 \right)}^{147}} + {{\left( 1 \right)}^{146}}.\left( { - 1} \right) + {{\left( 1 \right)}^{146}}}}{{{{\left( 1 \right)}^{145}}.\left( { - 1} \right) + {{\left( 1 \right)}^{145}} + {{\left( 1 \right)}^{144}}.\left( { - 1} \right) + {{\left( 1 \right)}^{144}} + {{\left( 1 \right)}^{143}}.\left( { - 1} \right)}} - 1\]
\[ \Rightarrow \dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = \dfrac{{1 - 1 + 1 - 1 + 1}}{{ - 1 + 1 - 1 + 1 - 1}} - 1\]By opening the bracket, we get,

By cancelling 1 with -1 from both numerator and denominator, we get,
\[ \Rightarrow \dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = \dfrac{1}{{ - 1}} - 1\]
\[ \Rightarrow \dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = - 1 - 1\]
\[ \Rightarrow \dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1 = - 2\]

Hence, the value of $\dfrac{{{i^{592}} + {i^{590}} + {i^{588}} + {i^{586}} + {i^{584}}}}{{{i^{582}} + {i^{580}} + {i^{578}} + {i^{576}} + {i^{574}}}} - 1$ is 2.

Note:
While solving this question students generally forget to put negative signs in ${i^2} = - 1$instead they put ${i^2} = 1$, which is wrong. It is noted that if the power of $i$ is even then the value of that numeric will be integer i.e. -1 or 1 but if the power of $i$ is odd then the value of that numeric will be a complex number i.e. $i$ or $ - i$. E.g.
1) ${i^{64}} = {i^{4 \times 16}} = 1$
2) ${i^{65}} = {i^{4 \times 16 + 1}} = {i^{4 \times 16}}.i = i$
3) ${i^{66}} = {i^{4 \times 16 + 2}} = {i^{4 \times 16}}.{i^2} = - 1$
4) ${i^{67}} = {i^{4 \times 16 + 3}} = {i^{4 \times 16}}.{i^3} = - i$