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Compute i^600+i^599+....+i+1
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The sequence of numbers
1, i, i^2, i^3, i^4, i^5, i^6, i^7, i^8, i^9, . . .
is periodic (cyclic) with the period of 4. Actually, it is
1, i, i^2, i^3, i^4=1, i^5=i, i^6=-1, i^7= -i, i^8=1, i^9=i, . . .
Why? Because i^4 = (i^2)^2 = (-1)^2 = 1.
So, your original sum is 150 times repeated the sum (1 + i + i^2 + i^3) PLUS the last addend i^600.
But i^2 = -1 and i^3 = -i, therefore the sum (1 + i + i^2 + i^3) is zero (!).
Hence, the original sum of 601 addends is equal to i^600 = 1.
Answer. i^600+i^599+....+i+1 = 1.
There is a bunch of my lessons on complex numbers
- Complex numbers and arithmetical operations on them
- Complex plane
- Addition and subtraction of complex numbers in complex plane
- Multiplication and division of complex numbers in complex plane
- Raising a complex number to an integer power
- How to take a root of a complex number
- Solution of the quadratic equation with real coefficients on complex domain
- How to take a square root of a complex number
- Solution of the quadratic equation with complex coefficients on complex domain
in this site.