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$x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$

What's the way to solve this kind of equation? I think there must be 8 solutions.

I tried to solve the following two equations

$a^6 = 1+i$

$b^6 = \sqrt{3} - i$

I tried the following for the first equation

$-1 = i^2 = (a^6 - 1)^2 = a^{12} - 2a^6 +1$

Then I would need to solve this:

$a^{12} - 2a^6 +2 = 0$

Is this the right approach?

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Hint: polar coordinates. –  GEdgar Dec 30 '11 at 17:27
    
Rewriting the term at the right in exponential form should help much! –  Raymond Manzoni Dec 30 '11 at 17:33
    
@RaymondManzoni: I began rewriting the term at the right. Is this what you mean? –  meinzlein Dec 30 '11 at 17:38
    
second hint : divide the numerator by sqrt(2) and the denominator by 2 (multiply the whole by sqrt(2) for compensation). You should recognize two well know exponentials. –  Raymond Manzoni Dec 30 '11 at 17:43
    
meinzlein: Not quite - the angles after the cos and sin should be the same, not different. For i+1, for example, think of it as a point on the complex plane. It's sqrt(2) away from the origin at an angle of pi/4, so i+1=sqrt(2)(cos(pi/4)+isin(pi/4)), which is sometimes abbreviated sqrt(2)cis(pi/4). Do you know De Moivre's Theorem for how to multiply and divide these polar forms? –  Lopsy Dec 30 '11 at 17:45
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4 Answers 4

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Any equation of the form $z^n=\alpha$, where $\alpha$ is a fixed complex number can be solved by switching $\alpha$ to polar form

$$\alpha= r (\cos(\theta)+i\sin(\theta) ) \,.$$

Now, can you find the polar forms of $z$?

For second equation, since the equation only has powers of $a^6$, a substitution $t=a^6$ should work ;)

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$(1+i)/\sqrt(2) = e^{i \pi/4}$
$(\sqrt(3)-i)/2= e^{-i \pi/6}$

You should obtain something like $z^8= a e^{i b}$ that should be solvable!

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You'll probably need De Moivre's Theorem to solve this nicely. De Moivre's Theorem states: If $R_1cis(w_1)$ and $R_2cis(w_2)$ are two complex numbers in polar form, then

$R_1cis(w_1)*R_2cis(w_2) = R_1R_2cis(w_1+w_2)$.

That is, to multiply two complex numbers in polar form, you multiply the distances and add the angles. You can use this to divide your numerator by the denominator (divide the distances and subtract the angles when dividing) to find the polar form for your RHS:

$\LARGE \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt{2}cis(\pi/4)}{2cis(\frac{-\pi}{6})} = \frac{1}{\sqrt{2}}cis(\frac{5\pi}{12})$.

Now, use the fact that $(Rcis(v))^8 = (R^8cis(8v))$ (eight applications of DeMoivre's) to find the polar form of x, and finally convert that back to a+bi form to find your answer! Don't forget: when you're dealing with angles, division isn't unique. For example, dividing 90 degrees by 2 can get you either 45 degrees or -135 degrees. You'll get eight different answers when you divide the angle by eight; hence, eight roots of the equation.

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Edited to change a number. Adding fractions is just too hard, apparently... –  Lopsy Dec 30 '11 at 18:22
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HINT:

$$x^8=\frac{1+i}{\sqrt{3}-i}=\frac{1+i}{\sqrt{3}-i}.\frac{\sqrt{3}+i}{\sqrt{3}+i}=\frac{\sqrt{3}-1}{4}+i\frac{1-\sqrt{3}}{4}$$

Then use polar coordinates, $x^8=r(cos\theta +i\sin\theta)$ to obtain all the 8 roots. Think of the identity $(r(cos\theta +i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$

(This is De-Moivre's Theorem)

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