What Are the Roots of the Equation ##z^4-2z^3+12z^2-14z+35=0##?

[Rectifier]

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.

Please help :,(

 

[Ray Vickson]

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.

Please help :,(

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]

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.

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.

Please help :,(

 

[ehild]

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.

Please help :,(

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

 

[SammyS]

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 .