# Homework Help: Complex Number agn.

1. Feb 21, 2010

### icystrike

1. The problem statement, all variables and given/known data
i)Find the fourth roots of the complex number -1+$$\sqrt{3}i$$ , giving your answer in the form of r$$e^{i\theta}$$ .

ii)Deduce the solution of the equation $$z^{8}+2z^{4}+4=0$$ , giving your answer exactly in the form r$$e^{i\theta}$$ .

2. Relevant equations

3. The attempt at a solution
i)$$2^{\frac{1}{4}}e^{(1+3k)\frac{\pi}{6}i}$$

s.t $$k=0,\pm1,2$$

ii)$$z^{4}=-1\pm\sqrt{3}i$$

z=$$2^{\frac{1}{4}}e^{(1+3k)\frac{\pi}{6}i}$$

or

z=$$2^{\frac{1}{4}}e^{(-1+3k)\frac{\pi}{6}i}$$

$$k=0,\pm1,2$$

They have only included the conjugate case.

Last edited: Feb 21, 2010
2. Feb 22, 2010

### Mentallic

Actually just a slight correction, it's $k=0, \pm1, -2$ since the restriction on $\theta$ is $-\pi<\theta\leq \pi$ and for k=2, the angle would be $7\pi/6>\pi$

The restriction on k for the conjugate case is correct.
Yep, they neglected the conjugate. It could just be a test to see if the students are able to realize they need 8 roots, and possibly even to be able to figure out the conjugate is required without solving for z in that polynomial just by inspection that the polynomial is real and if z is a complex root, then the conjugate is too.

Was that all that was troubling you?

3. Feb 22, 2010

### icystrike

yes. to summaries , there will be 8 roots ,the initial 4 roots and the 4 conjugate roots.
However, i thought the $$\theta$$ that is limited is referring to $$re^{i(\theta+2k\pi)}$$ , thus , whatever k is substituted does not matter so long as they are consecutive.

Last edited: Feb 22, 2010
4. Feb 22, 2010

### Mentallic

Actually it would be correct to say that k=0,1,2...(n-1) where n is the number of roots required because then you wouldn't have any multiple roots that clash. But take a look at your solutions: k=-1 and k=2 are the same root, and since you took k=-1 rather than just say k=0,1,2,3 then it is telling me that you are required to restrict $\theta$ to $-\pi<\theta\leq \pi$

And yes, of course the roots are going to be consecutive. but when you have an even number of roots to deal with - you must consider when you take $k=0,\pm1,\pm2...\pm(n-1)/2,x$ for even n roots, will $x=(n+1)/2$ or $-(n+1)/2$? Using actual numbers will make it easier to realize for what case x is positive or negative, but I can tell you now since I've learnt from my experience that if the principle root (the first root, with k=0) is positive, then x is negative and vice versa.

5. Feb 22, 2010

### tiny-tim

No, Mentallic is right …

look again at the question …
Just because you've decided to use the same letter, θ, as the question, that doesn't make it the same.

The whole exponent has to be the principal value.

(of course, if you really have an aversion to -2, you can write it e(3k - 2)π/6, k = 0, ±1, 2 )

6. Feb 22, 2010

### icystrike

Oh! got that! Thank you! (=