Complex solution of a cubic function.

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SUMMARY

The discussion focuses on solving the cubic equation z³ + iz² - 7z - iz - 6 - 6i = 0. The user successfully identified the two real solutions, 3 and -2, but struggled to find the complex solution. After some attempts, they concluded that the complex solution is -1 - i. The recommended method for finding the complex solution involves factoring the cubic equation by dividing it by the linear factors z + 2 and z - 3.

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yoyoyo992
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Hi,
I'm studying my algebra and I was trying this exercise:

Solve
z³+iz²-7z-iz-6-6i=0

I found the 2 real solutions (3 and -2) but i can't seem to find the complex one.

I tried this:

(z²-z)(i)=6(i) ==> z= -2 and 3
z³-7z=6==> z= -1, -2 and 3

I found that the complex solution is -1-i but i have no idea how to find it.

(Im from Belgium, it's difficult to type maths in another language so sorry for mistakes)
 
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Well, the obvious thing to do is divide by z+ 2 and z- 3 so that you can factor that cubic into linear terms, then set each term equal to 0.
 
Thanks, I found it
 

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