Question

Evaluate the following :

$\left(i\right)i^{457}\left(ii\right)i^{528}\left(iii\right)\frac{1}{i^{58}}\left(iv\right)i^{37}+\frac{1}{i^{47}}\left(v\right){(i^{41}+\frac{1}{i^{257}})}^9\left(vi\right)(i^{77}+i^{70}+i^{87}+i^{414}{)}^3(vii)i^{30}+i^{40}+i^{60}(viii)i^{49}+i^{68}+i^{89}+i^{110}$

Answer 1

$\left(i\right)i^{457}\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=1$

In order to find $i^n$ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

$n=4p+q,0\le q<4\text{Then}{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q=1^p\times i^q[\because 1^{p-1}]Hence{i}^n=i^q,where0\le q<4\therefore i^{457}=i^{4\times 114}\times i^1=i^1=i(ii)i^{528}\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=1$

In order to find $i^n$ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

$n=4p+q,0\le q<4Then{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q=1^p\times i^q[\because 1^{p-1}]Hence{i}^n=i^q,where0\le <q<4\therefore i^{528}=i^{4\times 32}=(i^4{)}^{132}=1^{132}=1\therefore \left(i^{528}\right)=1\left(iii\right)\frac{1}{i^{58}}\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=1$

In order to find $i^n$ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

$n=4p+q,0\le q<4\text{Then}{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q{1}^p\times i^q[\because 1p^{-1}]\text{Hence}{i}^n=i^q,\text{where}0\le q<4\therefore \frac{1}{i^{58}}=\frac{1}{i^{4\times 14}\times i^2}=\frac{1}{1\times i^2}=\frac{1}{-1}[\because i^2=-1]=-1(iv)i^{37}+\frac{1}{i^{47}}\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=-1$

In order to find $i^n$ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

$n=4p+q,0\le q<4\text{Then}{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q=1^p\times i^q[\because 1^{p-1}]\text{Hence}{i}^n=i^q,\text{where}0\le q<4\therefore i^{37}+\frac{1}{i^{67}}=i^{4\times 9}\times i^1+\frac{1}{i^{4\times 16}\times i^3}=1\times i^1+\frac{1}{1\times i^3}=i+\frac{1}{i^3\times i}\times i=i+\frac{1}{i^4}=i+\frac{i}{1}[\because i^4=1]=2i\therefore i^{37}+\frac{1}{i^{67}}=2i(V){(i^{41}+\frac{1}{i^{257}})}^9\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=1$

In order to find in where n > 4, divide n by 4 to get quotient p and remainder q, so that n = 4p + q, $0\le q<4$

$\text{Then}{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q=(1{)}^p\times i^q=i^q[\because 1^p=1]\text{Hence}{i}^n=i^q,\text{where}0\le q<4{[i^{41}+\frac{1}{i^{257}}]}^9=(i^{4\times 10}\times i^1+\frac{1}{i^{-4\times 64}\times i^1})={(1\times i+\frac{1}{1\times i})}^9={(i+\frac{1}{i})}^9={(i+\frac{1}{i\times 1}\times i)}^9={(i+\frac{i}{-1})}^9=(i-1{)}^9=0\therefore (i^{41}+\frac{1}{i^{257}})=0(vi\left)\right(i^{77}+i^{70}+i^{87}+i^{414}{)}^3\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=1$

In order to find $i^n$ where n > 4, we divide n by 4 get quotient p and remainder q, so that

$n=4p+q,0\le q<4\text{Then}{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q=1^p\times i^q[\because 1^{p-1}]=i^q\text{Hence}{i}^n=i^q,\text{where}0\le q<4(i^{77}+i^{70}+i^{87}+i^{414}{)}^3=(1\times i+1\times i^2+1\times i^3+1\times i^2{)}^3=(i-1-i-1{)}^3=(-2{)}^3=-8\therefore (i^{77}+i^{70}+i^{87}+i^{414}{)}^3=-8\left(vii\right)i^{30}+i^{40}+i^{60}\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=1$

In order to find $i^n$ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, $0\le q<4$

$\text{Then}{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q=i^p\times i^q=i^q[\because 1^{p-1}]\text{Hence}{i}^n=i^q,\text{where}0\le q<4\therefore i^{30}+i^{40}+i^{60}=i^{4\times 7}\times i^2+i^{4\times 10}+i^{4\times 15}=1\times i^2+1+1=-1+1+1=1\therefore i^{30}+i^{40}+i^{60}=1(viii)i^{49}+i^{68}+i^{89}+i^{110}\text{We know that}i=\sqrt{-1}{i}^2=-1i^3=-i{i}^4=1$

In order to find $i^n$ where n > 4, we divide n by 4 to get quotient p and remainder q, so that

n = 4p + q, $0\le q<4$

$\text{Then}{i}^n=i^{4p+q}=i^{4p}\times i^q=(i^4{)}^p\times i^q=1^p\times i^q=i^q[\because 1^{p-1}]\text{Hence}{i}^n=i^q,\text{where}0\le q<4\therefore i^{49}+i^{68}+i^{110}=i^{4\times 12}\times i^1+i^{4\times 17}+i^{4\times 22}\times i^{4\times 27}+i^2=1\times i+1+1\times i+1\times i^2=i+1+i-1=2i{i}^{49}+i^{68}+i^{89}+i^{110}=2i$