How do you simplify $i 44 + i 150 - i 74 - i 109 + i 61$ $i44 + i150 - i74 - i109 + i61$ ?

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How do you simplify $i 44 + i 150 - i 74 - i 109 + i 61$ $i44 + i150 - i74 - i109 + i61$ ?

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Answer 1 · 2016-03-01T18:50:15

1

Explanation:

We have,
$i^{1} = i$ $i^1=i$
$i^{2} = - 1$ $i^2= -1$
$i^{3} = i^{2} \times i = - i$ $i^3 = i^2 xx i = -i$
$i^{4} = i^{2} \times i^{2} = {(- 1)}^{2} \times {(- 1)}^{2} = 1$ $i^4 = i^2 xx i^2 =(-1)^2 xx (-1)^2 = 1$
$i^{5} = i^{4} \times i = 1 \times i = i$ $i^5 = i^4 xx i = 1 xx i =i$
$i^{6} = i^{4} \cdot i^{2} = 1 \cdot - 1 = - 1$ $i^6 = i^4 * i^2 = 1 * -1 = -1$
$i^{7} = i^{4} \cdot i^{3} = 1 \cdot - i = - i$ $i^7 = i^4 * i^3 = 1 * -i = -i$
$i^{8} = i^{4} \cdot i^{4} = 1 \cdot 1 = 1$ $i^8 = i^4 * i^4 = 1 * 1 = 1$

Now,
$i^{44} = {(i^{4})}^{11} = {(1)}^{11} = 1$ $i^44 = (i^4) ^11 = (1)^11 = 1$
$i^{150} = {(i^{4})}^{37} \cdot i^{2} = i^{2} = - 1$ $i^150 = (i^4)^37 * i^2 = i^2 = -1$
$i^{74} = {(i^{4})}^{18} \cdot i^{2} = i^{2} = - 1$ $i^74 = (i^4)^18 * i^2 = i^2 = -1$
$i^{109} = {(i^{4})}^{27} \cdot i^{1} = i$ $i^109 = (i^4)^27 * i^1 = i$
$i^{61} = {(i^{4})}^{15} \cdot i^{1} = i$ $i^61 = (i^4)^15 * i^1 = i$

Finally,
$i^{44} + i^{150} - i^{74} - i^{109} + i^{61}$ $i^44 + i^150 - i^74 - i^109 + i^61$
$= (1) + (- 1) - (- 1) - (i) + (i)$ $=(1) + (-1) - (-1) - (i) + (i)$
$= 1 - 1 + 1 - i + i$ $=1 - 1 + 1 - i + i$
$= 1$ $= 1$

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How do you simplify $i 44 + i 150 - i 74 - i 109 + i 61$ $i44 + i150 - i74 - i109 + i61$ ?

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How do you simplify i44+i150−i74−i109+i61i44 + i150 - i74 - i109 + i61?

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How do you simplify $i 44 + i 150 - i 74 - i 109 + i 61$ $i44 + i150 - i74 - i109 + i61$ ?