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I have an exam on this kind of stuff on Monday and was wondering if anyone could help me with these questions and can you tell me what this type of question is so i can go away and revise it for example(Further trig)

Question 1) The cube roots of 1 are denoted by $1, w$ and $w^2$ show that $1+w+w^2=0$

Question 2) The points A, B and C form an equilateral triangle they also represent the numbers $z_1,z_2,z_3$ State the geometrical effect of multiplication by $w$ and hence explain why: $z_1-z_3=w(z_3-z_2)$

Question 3) hence show that $z_1+z_2w+z_3w^2=0$

maxmitch
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5 Answers5

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Note that $$1+w+w^2=\frac{w^3-1}{w-1}$$

Of course, $w\neq 1$.

If you write $w=e^{i\theta}$, what is $\theta$? What does multiplication by $w$ then do? Remember that $e^{i\theta}e^{i\tau}=e^{i(\theta+\tau)}$.

Pedro
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Hints:

(1)

$$x^3-1=(x-1)(x^2+x+1)$$

The sum of all the roots of a polynomial of degree $\,n\,$ equals the polynomial's coefficient of $\,x^{n-1}\,$ divided by the leading coefficient.

(2)+(3) If

$$z=|z|e^{i\phi}\;,\;\;\text{then, since}\;\;w=e^{\frac{2\pi i}3}\,,\,\;\text{we get}:$$

$$wz=|z|e^{\left(\phi+\frac{2\pi}3\right)i}=|z|\left(\cos\left(\phi+\frac{2\pi}3\right)+i\sin\left(\phi+\frac{2\pi}3\right)\right)\implies$$

the number $\,wz\,$ has the same module as $\,z\,$ but it's rotated counterclockwise by an angle of $\,\frac{2\pi}3\,$ radians...

DonAntonio
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In general, the various roots of a monic polynomial add up to the negative of the subleading coefficient. The three cube roots you have all satisfy: $$x^3 - 1 =0$$ or rather $$x^3+\mathbf{0}x^2+0x-1=0$$ So this general fact would imply that $$1+\omega+\omega^2=-0$$

It seems to me that this problem should specify the orientation of the three points $z_1\to z_2\to z_3$ is counterclockwise.

Multiplication by $\omega$ rotates complex numbers by $120^\circ$ counterclockwise about $0$. Now $z_3-z_2$ represents a vector that represents one side of the triangle. Multiplying by $\omega$ would rotate $120^\circ$. So you would have a vector of the same length but rotated by $120^\circ$: exacly what you would get from $z_1-z_3$.

$$(z_1-z_3)+(z_2-z_1)+(z_3-z_2)=0$$ so $$(z_1-z_3)+\omega(z_1-z_3)+\omega(z_2-z_1)=0$$ and $$(z_1-z_3)+\omega^2(z_3-z_2)+\omega(z_2-z_1)=0$$ which implies $$(z_1+\omega z_2+\omega^2 z_3)(1-\omega)=0$$ and then we may divide by $(1-\omega)$.

2'5 9'2
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  • Sorry I forgot to mention that it is anti-clockwise but I have no idea what that means to this scenario! – maxmitch Jun 08 '13 at 20:05
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Hint: if you write the roots of unity as

$$ 1=e^{2\pi i}, \omega=e^{\frac{2\pi i}{3}}\implies \omega^2= e^{\frac{4\pi i}{3}}, $$

then things will be clear. To get the above roots, just note this

$$ z^3=1=e^{2k\pi i} \implies z= e^{\frac{2k\pi i}{3}}\quad k=0,1,2. $$

Mhenni Benghorbal
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Lots of answers are already here, however.....:

Let $\alpha = 1 + \omega + \omega^2$.

Then $\alpha\omega = \omega+\omega^2 +\omega^3$.

But $\omega^3=1$, so $\alpha\omega = \omega+\omega^2 +1$.

But this is the same as $\alpha$, so we have $\alpha\omega=\alpha$.

So $\alpha(\omega-1)=0$.

Thus either $\alpha=0$ (which is what we wanted to prove) or $\omega=1$.

So the thing to be proved is true if $\omega$ is either of the other cube roots of $1$ besides $1$ itself.

Michael Hardy
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