Determine Roots of Complex Polynomial

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• September 30th 2011, 03:09 AM
TaylorM0192
Determine Roots of Complex Polynomial
Hello,

I encountered this problem in my exercise set, and was stumped.

Solve (z + 1)^4 = 1 - i

My approach to the problem was to recognize that (1 - i) could be rewritten as [ (1 - i)^4 ] / 8.

Which then gives..

(z + 1)^4 = (1 - i)^4 / 8

(z + 1) = +/- (1 - i) / 4throot(8)

But I'm not sure how to proceed from here, since it would appear that in solving each of the resulting two equations, I would only obtain two roots, rather than the expected four.

Edit: Sorry, but LaTeX kept giving me an error when I tried using it.
• September 30th 2011, 03:34 AM
chisigma
Re: Determine Roots of Complex Polynomial
Quote:

Originally Posted by TaylorM0192
Hello,

I encountered this problem in my exercise set, and was stumped.

Solve (z + 1)^4 = 1 - i

My approach to the problem was to recognize that (1 - i) could be rewritten as [ (1 - i)^4 ] / 8.

Which then gives..

(z + 1)^4 = (1 - i)^4 / 8

(z + 1) = +/- (1 - i) / 4throot(8)

But I'm not sure how to proceed from here, since it would appear that in solving each of the resulting two equations, I would only obtain two roots, rather than the expected four.

Edit: Sorry, but LaTeX kept giving me an error when I tried using it.

Setting $s=z+1$ Your equation become...

$s^{4}= 2^{\frac{1}{2}}\ e^{i(2k-\frac{1}{4})\pi}$ (1)

... and its solution are...

$s= 2^{\frac{1}{8}}\ e^{i(k-\frac{1}{8}) \frac{\pi}{2}}$

... so that the solutions of the equation in z are...

$z= 2^{\frac{1}{8}}\ e^{i(k-\frac{1}{8}) \frac{\pi}{2}}-1$ (3)

Kind regards

$\chi$ $\sigma$
• September 30th 2011, 09:03 AM
TaylorM0192
Re: Determine Roots of Complex Polynomial
Hmm...I like this solution! Thank you!

Our text stipulates, however, that we should follow the "outline" used in one of the examples; i.e. with some clever algebraic manipulation put the complex polynomial equation in a solvable Cartesian form (i.e. do not use the polar form of the complex numbers). When I attempted it this way, I got stuck as I mentioned.

In any case, I think it's pretty dumb when in a few steps you can see (much more clearly at that) what the roots are.

I do have one question, though...is it acceptable to leave the solution in a non-standard form? i.e. the polar form with a "-1" tacked at the end? Or should you expand the polar form into its trigonometric representation and go through the grueling task of algebraically rewriting the expression in a standard form? If it is acceptable...then is it possible in this non-standard form to plot the roots (by inspection) on the complex plane, or is this pretty much an "analytic only" exact answer?

Thanks ~