Fourth Roots & Solutions of Complex # -1+$\sqrt{3}$i

[icystrike]

Homework Statement

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$$e^{i\theta}$$ .

Homework Equations

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.

 

[Mentallic]

The Attempt at a Solution

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

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

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 $7\pi/6>\pi$

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.

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?

 

[icystrike]

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?

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.

 

[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 learned from my experience that if the principle root (the first root, with k=0) is positive, then x is negative and vice versa.

 

[tiny-tim]

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.

No, Mentallic is right …

look again at the question …

… giving your answer exactly in the form r$$e^{i\theta}$$

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 )

 

[icystrike]

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 )

Oh! got that! Thank you! (=