**Complex Numbers - Cartesian Form**

**From WJEC FP1 (June 2005) - Question 1**

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The complex number *z* is represented by the point *P* on an Argand diagram.

Given that

*|* *z* + 1 | = 2 | *z* - 2i |

find, in its simplest form, the Cartesian equation of the locus of *P*. [5]

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Firstly *z* is a complex number and can be written in the form *x + iy**.* Since *z* is represented by the point *P* then the equation of the locus of *P* is an equation involving both *x* and *y*. So if we substitiute *x + iy* wherever there is a *z* then we have

| *x* + *iy* + 1 | = 2 | *x* + *iy* - 2*i* |

We must now collect the 'real' and 'imaginary' terms on each side together, which we will eventually square to remove the modulus on each side. Therefore

| (*x* + 1) + *iy* | = 2 | x + (*iy* - 2*i*) |

The definition of modulus in this instance is the square root of the sum of the real parts sqaured and the imaginary parts squared. So we have

√( (*x* + 1)² + *y*² ) = 2 √( *x*² + ( *y* - 2)² )

*(Note: We are taking the co-efficients of the imaginary parts here so *i*is not included)*

Expanding the brackets

√( (*x*² + 2*x* + 1) + *y*² ) = 2 √( *x*² + ( *y*² - 4*y* + 4) )

There is a square root on both sides of the equation, so I will square out both sides to remove them.

( *x*² + 2*x* + 1 ) + *y*² = 4 ( *x*² + ( *y*² - 4*y* + 4 ) )

Rearranging the right hand side by expanding out the brackets

*x*² + 2*x* + 1 + *y*² = 4*x*² + 4*y*² - 16*y* + 16

Collecting the like terms

0 = 4*x*² + 4*y*² - 16*y* + 16 - *x*² - 2*x* - 1 - *y*²

Then simplifying gives us

3*x*² + 3*y*² - 16*y* - 2*x* + 15 = 0

## Comments (2)

## Steph Richards said

at 5:58 pm on Dec 4, 2008

Good work Tom. Please tell me how you wrote the mathematical text.

## Tom Newton said

at 11:17 pm on Jan 8, 2009

The '²' is done by holding down the ALT key on your keyboard and pressing the code 0178 on the numpad. Likewise for '³' you use ALT+0179. Alternatively you could go to the 'Run' option on the Start menu of Windows and type 'charmap'. This opens up the list of all characters available for a particular font, and is how I got the square root sign.

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