How many solutions does the following equation have with positive real part?
z^4 + 8z^3 + 3z^2 + 8z + 3 = 0
I figured someone might have some fun with this problem since it requires so little knowledge to at least understand. I currently know of two solutions, but neither is obvious/elementary.
Here's a tiny bit of discussion to get things started. z here is a complex number; obviously if z is real and positive the left side is positive so there are no solutions in that case, so any potential solutions with positive real part must have non-zero imaginary part. From the fundamental theorem of calculus, there are 4 solutions (counted with multiplicity) overall. Since the coefficients are all real numbers, the solutions with non-zero imaginary part come in complex conjugate pairs. Thus there are 0, 2, or 4 solutions with positive real part. If you graph the function for real numbers, you find that it has precisely two solutions (both of course negative)--this fact can be justified rigorously without computation using some basic calculus and the intermediate value theorem. So, there are 0 or 2 solutions with positive real part. Which is it?
*The answer is (spoilerdontreadifyoudontwanttoknowtheansweristwothereisaidit).