How do you simplify i^48 + i^150 - i^78 - i^109 + i^61?

Dec 23, 2015

$1$. Is that a sentence?

Explanation:

Since $i = \sqrt{- 1} , {i}^{2} = - 1 ,$ so ${i}^{4} = {\left({i}^{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1.$

Look for multiples of 4 in the exponent:
$48 = 4 \cdot 12$, so ${i}^{48} = {\left({i}^{4}\right)}^{12} = {1}^{12} = 1$

$150 = 37 \cdot 4 + 2 ,$so ${i}^{150} = {\left({i}^{4}\right)}^{37} \cdot {i}^{2} = {1}^{37} \cdot \left(- 1\right) = - 1$
Similarly ${i}^{78} = - 1.$

$109 = 4 \cdot 54 + 1$, so ${i}^{109} = {1}^{54} \cdot i = i$
Similarly ${i}^{61} = i .$

${i}^{48} + {i}^{150} - {i}^{78} - {i}^{109} + {i}^{61} =$
$1 + \left(- 1\right) - \left(- 1\right) - i + i = 1.$