$\left(1-\frac{\sqrt{3}-i}{2}\right)^{24}$ somehow this should be equal to :$\left(2-\sqrt{3}\right)^{12}$ but I can't see how...
Need help to simplify the expression involving powers
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0@Michael Hardy: Yes, your'e right, that was stupid from my side :( – 2012-09-29
4 Answers
The easiest way to deal with imaginary numbers is with exponentials. This is definitely the case here. First, put everything over a common denominator, then take the exponential. Remember that $(e^{a})^{b} = e^{a*b}$ and $e^{a+b} = e^{a}*e^{b}.$ When you're done simplifying, take the ln of both sides. In the last step, think about what $(-i)^{12}$ equals.
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0Turns out this is because the two are the same except for a difference of (10^-19)*i (which may just be due to imprecision in my calculator). So for all intents and purposes they are the same. – 2012-09-29
Hint:
$ \left( 1 - \dfrac{\sqrt{3} - i}{2} \right)^{24} = \left[\left( 1 - \dfrac{\sqrt{3} - i}{2} \right)^2\right]^{12} $
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0@JavaMan : I've down-voted your answer since I don't see that it helps. I confess to a suspicion: You might have thought that if a certain complex number raised to the 12th power is $(2-\sqrt{3})^{12}$, then that complex number must be $2-\sqrt{3}$. Is that your whole rationale? If so, it's really easy to show that $\left(1-\dfrac{\sqrt{3}-i}{2}\right)^2$ is _not_ $2-\sqrt{3}$, and in fact it's not even a real number. – 2012-09-29
$ a=\frac{\sqrt{3}-i}{2} = \cos 30^\circ-i\sin30^\circ. $ Look at the triangle whose vertices are $0$, $a$, and $1$. Since the distance from $0$ to $1$ and the distance from $0$ to $a$ are both equal to the radius of the unit circle, the triangle is isosceles. The angle at the center of the circle is $30^\circ$ and the other two angles must be equal to each other. Since they have to add up to $180^\circ$, they must each be half of the remaining $150^\circ$, hence each $75^\circ$.
The short side of the triangle is just $1-a$. Hece $1-a=|1-a|(\cos75^\circ+i\sin75^\circ)$. Now $ |1-a|=\left|1-\frac{\sqrt{3}-i}{2}\right| = \left|\frac{2-\sqrt{3}-i}{2}\right| = \frac{2\sqrt{2-\sqrt{3}}}{2} = \sqrt{2-\sqrt{3}}. $
Hence $ (1-a)^{24} = \left(\sqrt{2-\sqrt{3}}\right)^{24} (\cos(24 \cdot 75^\circ) + i\sin(24 \cdot 75^\circ)) = \left(2-\sqrt{3}\right)^{12}\cdot(1). $ ($24\cdot75^\circ=1800^\circ = 5\text{ full circles}$, so the cosine is $1$ and the sine is $0$.)
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0Thankyou again. Really nice way you made it, man. – 2012-09-29
Hint: start by squaring $(1-\frac{\sqrt{3}-i}{2})$. Remember that $(a-b)^2 = a^2-2ab+b^2$, and that $i^2=-1$.
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0I suggest that _you_ try squaring $1-\frac{\sqrt{3}-i}{2}$ and _then_ tell us whether you think that helps. – 2012-09-29