# Complex numbers - factors

Gold Member
The problem
The following equation ##z^4-2z^3+12z^2-14z+35=0## has a root with the real component = 1. What are the other solutions?

The attempt
This means that solutions are ##z = 1 \pm bi##and the factors are ##(z-(1-bi))(z-(1+bi)) ## and thus ## (z-(1-bi))(z-(1+bi)) = z^2-z(1+bi)-z(1-bi)+(1-bi)(1+bi) = \\ = z^2 -z -zbi -z +zbi + 2 = z^2-2z+2 ##.

I tried to divide ##z^4-2z^3+12z^2-14z+35## with ## z^2-2z+2## but I get the quotient ##z^2 + 10## and rest ##6z + 15##

I have also tried to solve ##z^2-2z+2 = 0## and got ## z=1 \pm i ##, thus ## b = \pm 1 ##. But that is wrong according to key in my book.

Ray Vickson
Homework Helper
Dearly Missed
The problem
The following equation ##z^4-2z^3+12z^2-14z+35=0## has a root with the real component = 1. What are the other solutions?

The attempt
This means that solutions are ##z = 1 \pm bi##and the factors are ##(z-(1-bi))(z-(1+bi)) ## and thus ## (z-(1-bi))(z-(1+bi)) = z^2-z(1+bi)-z(1-bi)+(1-bi)(1+bi) = \\ = z^2 -z -zbi -z +zbi + 2 = z^2-2z+2 ##.

I tried to divide ##z^4-2z^3+12z^2-14z+35## with ## z^2-2z+2## but I get the quotient ##z^2 + 10## and rest ##6z + 15##

I have also tried to solve ##z^2-2z+2 = 0## and got ## z=1 \pm i ##, thus ## b = \pm 1 ##. But that is wrong according to key in my book.

Substitute ##z = 1+i b## into the original equation. You will get a complex 4th degree polyomial in the new variable ##b##. Equating its real and imaginary parts to zero (assuming ##b## is real) you get two equations for ##b##, and one of them is easy to solve.

Mark44
Mentor
The problem
The following equation ##z^4-2z^3+12z^2-14z+35=0## has a root with the real component = 1. What are the other solutions?

The attempt
This means that solutions are ##z = 1 \pm bi##and the factors are ##(z-(1-bi))(z-(1+bi)) ## and thus ## (z-(1-bi))(z-(1+bi)) = z^2-z(1+bi)-z(1-bi)+(1-bi)(1+bi) = \\ = z^2 -z -zbi -z +zbi + 2 = z^2-2z+2 ##.
Your mistake is above. (1 + bi)(1 - bi) ≠ 2.
Rectifier said:
I tried to divide ##z^4-2z^3+12z^2-14z+35## with ## z^2-2z+2## but I get the quotient ##z^2 + 10## and rest ##6z + 15##

I have also tried to solve ##z^2-2z+2 = 0## and got ## z=1 \pm i ##, thus ## b = \pm 1 ##. But that is wrong according to key in my book.

ehild
Homework Helper
The problem
The following equation ##z^4-2z^3+12z^2-14z+35=0## has a root with the real component = 1. What are the other solutions?

The attempt
This means that solutions are ##z = 1 \pm bi##and the factors are ##(z-(1-bi))(z-(1+bi)) ## and thus ## (z-(1-bi))(z-(1+bi)) = z^2-z(1+bi)-z(1-bi)+(1-bi)(1+bi) = \\ = z^2 -z -zbi -z +zbi + 2 = z^2-2z+2 ##.

This is wrong. Try again.

I tried to divide ##z^4-2z^3+12z^2-14z+35## with ## z^2-2z+2## but I get the quotient ##z^2 + 10## and rest ##6z + 15##

I have also tried to solve ##z^2-2z+2 = 0## and got ## z=1 \pm i ##, thus ## b = \pm 1 ##. But that is wrong according to key in my book.

Your method will work if you divide by the correct product of (z-1+bi)(z-1-bi). It must contain b!

SammyS
Staff Emeritus
Homework Helper
Gold Member
It may help to rewrite (z−1+bi)(z−1−bi) as:
##\displaystyle \left((z-1)+bi \right)\left((z-1)-bi \right)\,, ##​
which is a sum times a difference, thus the difference of squares .