# Help solving a complex equation

1. May 12, 2005

### Yura

im not sure i can fully remember the rules for complex numbers but i have to solve an equation that has 3 solutions to equal zero. so far i have (x=2), (x=i) and i think there was a + rule for a solution with complex numbers that there would always be a conjugate solution of it. so i figured that the last solution would be (x=-i) but when i try and solve it i keep getting different answers and they are all non zero.

i think i may be doing something wrong with the powers of the (-i) or it might be that i thought wrong and (x=-i) is not a solution.

could someone please confirm ths for me?

here is the equation:
x^3 - (2-i)*x^2 + (2-2*i)*x - 4 = 0

im trying to solve for:
(-i)^3 - (2-i)*(-i)^2 + (2-2*i)*(-i) - 4 = 0

thanks in advance ^^

2. May 12, 2005

### dextercioby

No,it doesn't have real coefficients.

Dividing the polynomial

$$x^{3}-(2-i)x^{2}+(2-2i)x-4$$ by $$(x-2)$$

u get the polynomial $$x^{2}+ix+2$$ which has the solutions $$x_{1}=i$$ and $$x_{2}=-2i$$ .

Daniel.

3. May 12, 2005

### arildno

You really shouldn't try to merely remember rules, you should understand them in the first place!
You cannot use the "conjugate rule" here (when does that one apply?)

To give you a hint:
Use polynomial division to determine the last root.

EDIT:
Daniel is a real polynomial divisionist, a rather complex person, actually.
He doesn't need reminders in order to find remainders..

Last edited: May 12, 2005
4. May 12, 2005

### Yura

ah ok thankyou ( i had actually tried polynomial long division first but half way through i remembered my one of my teachers saying something about a conjugate rule for it so i stopped and tried that first) now that i try the division i find it works out fine, so i'll have to look up that rule and see if it works or not

thanks again

Last edited: May 12, 2005
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