Author Topic: Complex Numbers Help! (Read 703 times) Tweet Share

Edward Elric · « **on:** June 07, 2014, 03:33:50 pm »

Given that 1-i is a solution to the equation

z^4 - 4Z^3 + 9Z^2 -10z +6 = 0

find the other solutions.

b^3 · « **Reply #1 on:** June 07, 2014, 03:58:15 pm »

As the coefficients are all real, we can use the conjugate theorem to tell us that $z-(1+i)$ is also a factor.
$\begin{alignedat}{1}\left(z-\left(1-i\right)\right)\left(z-\left(1+i\right)\right) & =z^{2}-z\left(1-i+1+i\right)+\left(1-i\right)\left(1+i\right) \\ & =z^{2}-2z+\left(1-i^{2}\right) \\ & =z^{2}-2z+2 \end{alignedat}$

Now long dividing through gives:
$\begin{alignedat}{1} & \underline{\; z^{2}-2z+3\qquad\qquad\qquad} \\ z^{2}-2z+2 & |z^{4}-4z^{3}+9z^{2}-10z+6 \\ \left(-\right) & \underline{\; z^{4}-2z^{3}+2z^{2}\qquad} \\ & \quad\quad-2z^{3}+7z^{2}-10z+6 \\ \left(-\right) & \quad\quad\underline{-2z^{3}+4z^{2}-4z\qquad} \\ & \qquad\qquad\quad\:3z^{2}-6z+6 \\ \left(-\right) & \qquad\qquad\quad\:\underline{3z^{2}-6z+6\qquad} \\ & \qquad\qquad\qquad\qquad\qquad0 \\ \implies z^{4}-4z^{3}+9z^{2}-10z+6&=\left(z^{2}-2z+2\right)\left(z^{2}-2z+3\right) \end{alignedat}$

This leaves us with factorising the left over part.
$\begin{alignedat}{1}z^{2}-2z+3 & =0 \\ z^{2}-2z+\left(1\right)^{2}-\left(1\right)^{2}+3 & =0 \\ \left(z-1\right)^{2}-2i^{2} & =0 \\ \left(z-1-\sqrt{2}i\right)\left(z-1+\sqrt{2}i\right) & =0 \\ \implies z & =1\pm\sqrt{2}i \\ \therefore z & =1-i,\:1+i,\:1-\sqrt{2}i,\:1+\sqrt{2}i \end{alignedat}$

Edward Elric · « **Reply #2 on:** June 07, 2014, 04:19:22 pm »

Quote from: b^3 on June 07, 2014, 03:58:15 pm

As the coefficients are all real, we can use the conjugate theorem to tell us that $z-(1+i)$ is also a factor.
$\begin{alignedat}{1}\left(z-\left(1-i\right)\right)\left(z-\left(1+i\right)\right) & =z^{2}-z\left(1-i+1+i\right)+\left(1-i\right)\left(1+i\right) \\ & =z^{2}-2z+\left(1-i^{2}\right) \\ & =z^{2}-2z+2 \end{alignedat}$

Now long dividing through gives:
$\begin{alignedat}{1} & \underline{\; z^{2}-2z+3\qquad\qquad\qquad} \\ z^{2}-2z+2 & |z^{4}-4z^{3}+9z^{2}-10z+6 \\ \left(-\right) & \underline{\; z^{4}-2z^{3}+2z^{2}\qquad} \\ & \quad\quad-2z^{3}+7z^{2}-10z+6 \\ \left(-\right) & \quad\quad\underline{-2z^{3}+4z^{2}-4z\qquad} \\ & \qquad\qquad\quad\:3z^{2}-6z+6 \\ \left(-\right) & \qquad\qquad\quad\:\underline{3z^{2}-6z+6\qquad} \\ & \qquad\qquad\qquad\qquad\qquad0 \\ \implies z^{4}-4z^{3}+9z^{2}-10z+6&=\left(z^{2}-2z+2\right)\left(z^{2}-2z+3\right) \end{alignedat}$

This leaves us with factorising the left over part.
$\begin{alignedat}{1}z^{2}-2z+3 & =0 \\ z^{2}-2z+\left(1\right)^{2}-\left(1\right)^{2}+3 & =0 \\ \left(z-1\right)^{2}-2i^{2} & =0 \\ \left(z-1-\sqrt{2}i\right)\left(z-1+\sqrt{2}i\right) & =0 \\ \implies z & =1\pm\sqrt{2}i \\ \therefore z & =1-i,\:1+i,\:1-\sqrt{2}i,\:1+\sqrt{2}i \end{alignedat}$

Thank You Very Much

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Author Topic: Complex Numbers Help! (Read 703 times) Tweet Share

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Complex Numbers Help!

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