- #1

Rectifier

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**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 :,(

Please help :,(