# Need help to simplify the expression involving powers

$$\left(1-\frac{\sqrt{3}-i}{2}\right)^{24}$$ somehow this should be equal to :$$\left(2-\sqrt{3}\right)^{12}$$ but I can't see how...

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Unequal complex numbers can still have equal $12$th powers. –  GEdgar Sep 29 '12 at 17:39
The argument of $b=1-(\sqrt{3}-i)/2$ is $(5/12)\pi$, so $b^{24}$ is a positive real number. Then compute the absolute value. –  GEdgar Sep 29 '12 at 17:44
Well, $b$ is in the first quadrant, and $\arctan((1/2)/(1-\sqrt3/2))/\pi$ came out $0.416666666$ so I was pretty sure it was $5/12$. –  GEdgar Sep 29 '12 at 17:59
All this commentary and I'm the only one who's upvoted the question! (I'm also the only one who's posted a correct answer....) –  Michael Hardy Sep 29 '12 at 18:16
@Michael Hardy: Yes, your'e right, that was stupid from my side :( –  Dennis Gulko Sep 29 '12 at 18:17

$$a=\frac{\sqrt{3}-i}{2} = \cos 30^\circ-i\sin30^\circ.$$ 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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(Recall that if $a$, $b$ are real, then $|a+bi|=\sqrt{a^2+b^2}$. I used that.) –  Michael Hardy Sep 29 '12 at 17:59
Of course, an alternative would be to expand via the binomial theorem and then simplify, but that's rather onerous. –  Michael Hardy Sep 29 '12 at 18:02
Thankyou @Michael Hardy. Nice way. I am just worried about that -1 –  Mykolas Sep 29 '12 at 18:11
@Mykolas : The $-1$ was there because I mistakenly multiplied by 12 instead of 24. I've fixed that now. So could you consider up-voting and possibly "accepting" this answer? –  Michael Hardy Sep 29 '12 at 18:13
Thankyou again. Really nice way you made it, man. –  Mykolas Sep 29 '12 at 18:13

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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I get $\frac{(3-2\sqrt{3})+(2-\sqrt{3})i}{2}$ for the square of the expretion, but I can't see how this helps me...@JavaMan could you explain bit more? –  Mykolas Sep 29 '12 at 17:23
I also don't see that it helps. But I've posted a separate answer below. –  Michael Hardy Sep 29 '12 at 17:53
@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. –  Michael Hardy Sep 29 '12 at 18:09

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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I suggest that you try squaring $1-\frac{\sqrt{3}-i}{2}$ and then tell us whether you think that helps. –  Michael Hardy Sep 29 '12 at 18:10
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.
$(e^a)^b=e^{(ab)}$ but $e^{a^b}$ means $e^{(a^b)}$, not $(e^a)^b$. And $e^{ab}$ is certainly not the same as $e^a\cdot e^b$, which is equal to $e^{a+b}$. –  Michael Hardy Sep 29 '12 at 17:36