Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.6 - Complex Numbers - 1.6 Exercises - Page 64: 86



Work Step by Step

We know that any number to the power of zero is 1, so $i^0=1$ We also know that $i^2=-1$, so $i^3=i^2 \cdot i^1=-1 \cdot i=-i$ and $i^4=i^2\cdot i^2=-1\cdot -1=1$ Using this iteration process, we'll can find that: $i^5=i^4\cdot i^1=1\cdot i=i$ $i^6=i^4\cdot i^2=1\cdot -1=-1$ $i^7=i^4\cdot i^3=1\cdot -i=-i$ $i^8=i^4\cdot i^4=1\cdot 1=1$ $i^9=i^4\cdot i^4 \cdot i^1=1\cdot 1 \cdot i=i$ $i^{10}=i^4\cdot i^4 \cdot i^2=1\cdot 1\cdot -1=-1$ $i^{11}=i^4\cdot i^4 \cdot i^3=1\cdot 1\cdot -i=-i$ $i^{12}=i^4\cdot i^4 \cdot i^4=1\cdot 1\cdot 1=1$ Putting them all up in order, we see that the pattern $1, i, -1, -i$ repeats itself: $i^0=1$ $i^1=i$ $i^2=-1$ $i^3=-i$ $i^4=1$ $i^5=i$ $i^6=-1$ $i^7=-1$ $i^8=1$ $i^9=i$ $i^{10}=-1$ $i^{11}=-1$ $i^{12}=1$ We can see in the iteration process that any power of $i$ can be simplified to $i^0, i^1, i^2,$ or $i^3$ by reducing the exponent by 4 as many times as needed since $i^4 =1$ and all numbers multiplied by $1$ gives that same number. Therefore to calculate any power of $i$, we'll need to divide the exponent by 4 and find the remainder. The solution will be $i$ to the power of the remainder. It works well because the remainder can only be 0, 1, 2 or 3. Now we can calculate $i^{4446}$. First, we divide 4446 by 4, giving us 1111 and a remainder of 2. So, the answer is $i^2$ which is $-1$.
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