# Complex Number problems.

Heyas. I'm not too good at complex numbers so excuse me if these questions are a bit on the laughable side.

Find all real values for r for which ri is a solution of the equation.
z^4 - 2z^3 + 11z^2 - 18z + 18 = 0
hence, Determine all the solutions of the equations...

I'm not really sure how to tackle questions like this. I subbed in 'ri' for 'z' and worked out r, but it had multiple values, some of which are not solutiuons as listed in the back of the book. Perhaps i did it wrong all together.

A quick guide through the question would help heaps :)

Also

A similar question
Find the real number 'k' such that z=ki is a root of the equation:
z^3 + (2+i)z^2 + (2+2i)z + 4 = 0.

is this basically the same as the earlier question?

HallsofIvy
Homework Helper
Okay, assuming you are given that there is an imaginary solution,
replace z with ri. Then, as I am sure you calculated, z^2= -r^2,
z^3= -r^3 i and z^4= r^4. Substituting those into the equation,
r^4+ 2r^3i- 11r^2- 18ri+ 18= 0.

Of course, in order for this to be 0, both real and imaginary parts must be 0. The real part is r^4- 11r^2+ 18= 0 and the imaginary part is 2r^3- 18r= 0. The crucial part is that the same r must satisfy BOTH equations. It's easy to factor the second equation:
2r^3- 18r= 2r(r^2- 9)= 0. The solutions are r=0, r= 3 and r= -3. It's obvious that r=0 does not satisfy r^4- 11r^2+ 18= 0. What about r= 3 and r= -3?

Now, for z^3 + (2+i)z^2 + (2+2i)z + 4 = 0.

Again, if z= ri, then z^2= -r^2 and z^3= -r^3 i. Substituting that into the equation, -r^3 i+ (2+i)(-r^2)+ (2+ 2i)(ri)+ 4=
-r^3 i- 2r^2- r^2 i+ 2ri- 2r+ 4= 0.

Separate real and imaginary parts: the real part is
-2r^2- 2r+ 4= 0. The imaginary part is -r^3- r^2+ 2r= 0.
The first of those factors as -2(r+2)(r-1)= 0 and so has solutions r= 1, r= -2. The second factors as -r(r^2+ r- 2)= -r(r+2)(r-1)= 0 and so has solutions r= 0, r= 1, r= -2. Remembering that r must satisfy BOTH equations, what can r be?

Thanks for that man :)