Author Topic: Finding solutions of complex equations (Read 1318 times) Tweet Share

squance · « **on:** September 18, 2008, 08:31:35 pm »

Find all the solutions of the following equation.

z^4 -5z^2 -8z - 12 =0

I don't know how to do this one but the answerss are 3, -2, -1/2 + (i sqrt7)/
2, -1/2 - (i sqrt7)/2

Some help would be appreciated!

ell · « **Reply #1 on:** September 18, 2008, 09:39:23 pm »

Hmm - there's probably a better and easier way than this, but you could let $P(z)=z^4 -5z^2 -8z - 12$ , then find when $P(z)=0$ by substituting values like 1, 2, 3 and -1, -2, -3 etc.. (works out that $z-3$ and $z+2$ are factors)

Then use long division and divide $z^4 -5z^2 -8z - 12$ by $z-3$ to obtain $P(z)=(z-3)(z^3+3z^2+4z+4)$

Then, using long division again, divide $z^3+3z^2+4z+4$ by $z+2$ to obtain $P(z)=(z-3)(z+2)(z^2+z+2)$

You can complete the square on the quadratic:

$z^2+z+2$

$\implies z^2+z+\frac{1}{4}+\frac{7}{4}$

$\implies \left(z+\frac{1}{2}\right)^2+\frac{7}{4}$

$\implies \left(z+\frac{1}{2}\right)^2-\frac{7 i^{2}}{4}$

$\implies \left(z+\frac{1}{2}- \frac{i\sqrt{7}}{2}\right)\left(z+\frac{1}{2}+ \frac{i\sqrt{7}}{2}\right)$

Which means:

$(z-3)(z+2)\left(z+\frac{1}{2}- \frac{i\sqrt{7}}{2}\right)\left(z+\frac{1}{2}+ \frac{i\sqrt{7}}{2}\right)=0$

Then find the solutions for z from there.

squance · « **Reply #2 on:** September 19, 2008, 09:50:18 am »

Thanks ell!

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Author Topic: Finding solutions of complex equations (Read 1318 times) Tweet Share

squance

Finding solutions of complex equations

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