How do you simplify $3i^42 - 5i^46 + 2i^29 - (4i^2)^3 + i^0$ ?

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Answer 1 · 2015-12-06T15:15:09

$3i^42-5i^46+2i^29-(4i^2)^3+i^0 = 67+2i$

For any integer $k$ :

$i^(4k+0) = 1$

$i^(4k+1) = i$

$i^(4k+2) = -1$

$i^(4k+3) = -i$

So:

$3i^42-5i^46+2i^29-(4i^2)^3+i^0$

$=3(-1)-5(-1)+2(i)-4^3(-1)+1$

$=-3+5+2i+64+1 = 67+2i$

How do you simplify 3i^42 - 5i^46 + 2i^29 - (4i^2)^3 + i^03i^42 - 5i^46 + 2i^29 - (4i^2)^3 + i^0?