### Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

### 8 Methods for Three by One

This problem in geometry has been solved in no less than EIGHT ways by a pair of students. How would you solve it? How many of their solutions can you follow? How are they the same or different? Which do you like best?

### An Introduction to Complex Numbers

A short introduction to complex numbers written primarily for students aged 14 to 19.

# Target Six

##### Stage: 5 Challenge Level:

This solution is from Arun Iyer, SIA High School and Junior College.

The given equation is $x^{3/2} - 8x^{-3/2} = 7$.

Multiply throughout by $x^{3/2}$and rearranging we get

$(x^{3/2})^2 - 7x^{3/2} - 8 = 0$

This can be factorised as:

$(x^{3/2} - 8)(x^{3/2} + 1) = 0$

Case 1:

Consider,

$x^{3/2} + 1 = 0$ or $x^{3/2} = -1$

Squaring both sides, $x^3 = 1$ so the three cube roots of unity are solutions.

Now

$1 = e^{2k\pi i}$

(for $k=0,1,2,3$...) therefore $x = e^{{2k\pi i} /3}$ .

Putting $k=0$, $1$, $2$ we get:

$x = e^0$ (which is $1$) or

$x = exp({\pm({2\pi i} /3)})$

.

Case 2:

Consider,

$x^{3/2}- 8 = 0$ or $x^{3/2} = 8$

Squaring both sides, $x^3 = 64$ so the three cube roots of $64$ are solutions.

Now

$x^3 = 64 e^{2k\pi i}$

(for $k=0,1,2,3$...) therefore

$x = 64^{1/3} e^{{2k\pi i}/3}$ .

Putting $k=0,1,2$ we get: $x = 4$ or

$x = 4exp({\pm{2\pi i} /3})$

.

So the six roots are:

• $x_1 = 1$
• $x_2 = \exp({2\pi i} /3) = \cos (2\pi /3) + i\sin(2\pi /3) = -1/2 + i {\sqrt3}/2$
• $x_3 = \exp({-2\pi i} /3) = \cos (-2\pi /3) + i\sin(-2\pi /3) = -1/2 - i {\sqrt3}/2$
• $x_4 = 4$
• $x_5 = 4\exp({2\pi i} /3) = 4\cos (2\pi /3) + 4i\sin(2\pi /3) = -2 + 2i {\sqrt3}$
• $x_6 = 4\exp({-2\pi i} /3) = 4\cos (-2\pi /3) + 4i\sin(-2\pi /3) = -2 - 2i {\sqrt3}$

Substituting $x=1$ into the given equation we need to recognise that $1^{3/2}$ has two values $+1$ and $-1$ so that whereas $x^{3/2} = 1$ does not satisfy the equation the other value $x^{3/2} = -1$ does satisfy it and hence $x=1$ is a solution of the equation.