Off the bat,
y = sqrt(i*y)
y^2 = i*y
y(y-i) = 0
y = {0, i}
Conceptually, in the complex plane if z has magnitude (or modulus) of 1, then multiplying by i rotates its position 90 degrees counterclockwise; taking the square root rotates it clockwise half the distance toward zero. So y=i makes sense. And y=0 makes sense since rotating a vector of zero modulus keeps it in the same place.
But what if we go further:
y=sqrt(i*sqrt(i*y))
y^2 = i*sqrt(i*y)
y^4 = (-1)(i*y)
y(y^3 + i) = 0
y(y^3 + (-i)^3) = 0
y(y^3 - i^3) = 0
y(y-i)(y^2 + i*y + i^2)
y(y-i)(y^2 + i*y - 1)
y = {0, i, (√3 - i)/2, (-√3 - i)/2}
or 0 and the 3 cube roots of -i
So, I'm guessing that if you go another step, then square both sides 3 times, you might get a 6th degree polynomial whose solutions might be 0 and i and maybe the other 4 fifth roots of i.
This could be extended to infinite solutions. Or could it?
In the end, I think all those other solutions besides 0 and i are probably extraneous solutions and don't satisfy y = sqrt(i*y)
y = {0, i} final answer.
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Posted by Larry
on 2023-04-23 10:16:56 |
my answer
