Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
3  
Unequal complex numbers can still have equal $12$th powers. –  GEdgar Sep 29 '12 at 17:39
1  
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
add comment

4 Answers

up vote 1 down vote accepted

$$ a=\frac{\sqrt{3}-i}{2} = \cos 30^\circ-i\sin30^\circ. $$ 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 $0$.)

share|improve this answer
    
(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
add comment

Hint:

$$ \left( 1 - \dfrac{\sqrt{3} - i}{2} \right)^{24} = \left[\left( 1 - \dfrac{\sqrt{3} - i}{2} \right)^2\right]^{12} $$

share|improve this answer
1  
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
add comment

Hint: start by squaring $(1-\frac{\sqrt{3}-i}{2})$. Remember that $(a-b)^2 = a^2-2ab+b^2$, and that $i^2=-1$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
$(e^a)^b=e^{(ab)}$ 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
    
WolframAlpha gives the same alternate form for the two given formulas. Both equal 3650401 - 2107560*sqrt(3). –  Andy Harris Sep 29 '12 at 17:55
    
Turns 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. –  Andy Harris Sep 29 '12 at 18:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.