Complex Numbers Problem (1 Viewer)

[ShawnG]

Can someone help me solve this?

z^4 - (z + 4)^4 = 0

I got +1.633i , - 1.633i and -6 , whilst my friend claims to have gotten 2+2i and -2-2i.
Thanks

 

[alakazimmy]

Can someone help me solve this?

z^4 - (z + 4)^4 = 0

I got +1.633i , - 1.633i and -6 , whilst my friend claims to have gotten 2+2i and -2-2i.
Thanks

Use difference between two squares:

(z^2-(z+4)^2)*(z^2+(z+4)^2)=0

(z- (z+4))*(z+(z+4))*(z^2+(z+4)^2)=0

For the first 2 factors, simplifying, you'll get: -4*(2z+4)
Obviously, -4 can't be 0, so 2z=-4, hence z = -2 is one solution

For the 3rd factor, use difference between 2 squares again.
+(z+4)^2 is the same as -(i(z+4))^2.

Hence, (z^2+(z+4)^2) = (z-i(z+4))*(z+i(z+4))
= (z-iz-4i)*(z+iz+4i)

So z-iz-4i = 0, or z+iz+4i = 0

Now z-iz-4i = z(1-i)-4i
So z = 4i/(1-i) = -2 + 2i

And z+iz+4i = z(1+i)+4i
So z= -4i/(1+i) = -2 - 2i

So the 3 solutions are: -2, -2+2i and -2-2i

 

[alakazimmy]

Also, to verify your answer, you can use the conjugate root rule.
All coefficients of the original question are real, so the complex roots must be conjugates, and they are.