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TOM NEWTON

Page history last edited by Tom Newton 15 years, 3 months ago

 

Complex Numbers - Cartesian Form

 

From WJEC FP1 (June 2005) - Question 1

 

_______________________________________________________________

 

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]

_______________________________________________________________

 

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 - 2i |

 

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 - 2i) |

 

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 + y² ) = 2 √( x² + ( y - 2)² )

 

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

 

Expanding the brackets

 

                                             √( (x² + 2x + 1) + y² ) = 2 √( x² + ( y² - 4y + 4) )

 

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

 

                                             ( x² + 2x + 1 ) + y² = 4 ( x² + ( y² - 4y + 4 ) )

 

Rearranging the right hand side by expanding out the brackets

 

                                             x² + 2x + 1 + y² = 4x² + 4y² - 16y + 16

 

Collecting the like terms

 

                                             0 = 4x² + 4y² - 16y + 16 - x² - 2x - 1 - y²

 

Then simplifying gives us

 

                                             3x² + 3y² - 16y - 2x + 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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