I was given this task: $$z^4+4z+8=0$$
And I am totally stuck at solving this. The university handed out some solutions that are the following:$$z=−2±2i$$
And well... plugging this into the function obviously works but I have no idea on how to get to that solution.
What I tried was to seperate the term into two halfs that look like this: $$z^4+4z+8=(z+a)^2\cdot(z+b)^2 = 0$$ Then I could have solved $$(z+a)^2 = 0$$ $$(z+b)^2 = 0 $$
But I did not find an a and b that work in this case. Therefor I am asking how this would usually be solved.
Greetings, Finn
I think the maximum, which you can make here it's the following. $$z^4+4z+8=(z^2+k)^2-2kz^2-k^2+4z+8=(z^2+k)^2-(2kz^2-4z+k^2-8).$$ Now, choose a value of $k$ such that $$2^2-2k(k^2-8)=0$$ or $$k^3-8k-2=0.$$ One of real roots of the last equation will get the difference of squares
and you'll get two quadratic equations and all roots, but it's very ugly of course.