complex question (1 Viewer)

[hatty]

hey there


P = 3 - 2i
Q = -3 + 2i

A point R may be found in the 1st quadrant so that the triangle PQR is equilateral. Show that R represents the complex number

sqrt.3 (2 + 3i )

that reads as "the square root of 3 * (2 + 3 i ) "

thanks for all assistance

 

[wogboy]

Draw up the triangle PQR on an Argand diagram.

The vector PR is simply the vector PQ rotated pi/3 radians clockwise (since angle QPR is 60 degrees, and lengths PR = PQ, for a equilateral triangle).

PQ
=Q - P
= (-3 + 2i) - (3 - 2i)
= - 6 + 4i

to rotate by pi/3 clockwise, multiply by:
[cos(-pi/3) + isin(-pi/3)]
= 1/2 - i*sqrt(3)/2

so PR = (- 6 + 4i)(1/2 - i*sqrt(3)/2)
= -3 + 2*sqrt(3) + i*(3*sqrt(3) + 2)

now PR = R - P
R = PR + P
= 2*sqrt(3) + i*[3*sqrt(3)]
= sqrt(3)*[2 + 3i] (as required)

 

[CM_Tutor]

Wogboy's vector method is the best way to do this, but if you have trouble with geometric methods, there is an algebraic alternative, which is:

- Find the distance PQ
- Let the required point R be x + iy
- Since its an equilateral triangle, PR = QR = PQ, so form two simultaneous equations in the unknowns x and y from PR = PQ and QR = PQ
- Solve for x and y - there will be two answers - and discount the one where x and / or y are negative.
- you should be left with x = 2 * sqrt(3) and y = 3 * sqrt(3), and so you have R is at sqrt(3) * (2 + 3i) as required.

 

[hatty]

thanks guys

wogboy i dont understand ur method

PQ
=Q - P
= (-3 + 2i) - (3 - 2i)
= - 6 + 4i

why does PQ = Q - P ?

cheers

 

[hatty]

cm tutor

could u do this step for me please

Since its an equilateral triangle, PR = QR = PQ, so form two simultaneous equations in the unknowns x and y from PR = PQ and QR = PQ


thanks

 

[:: ck ::]

Q - P

is how we define the vector FROM P to Q

 

[CM_Tutor]

The distance PQ is sqrt(52). So...

PR = PQ
PR^2 = PQ^2
PR^2 = 52
Now, P represents the complex number 3 - 2i, and R represents x + iy.
So PR^2 = (x - 3)^2 + (y + 2)^2 = 52 _______ (1)

Similarly, given Q represents -3 + 2i,
QR^2 = (x + 3)^2 + (y - 2)^2 = 52 _______ (2)

Now (1) = (2), so (x - 3)^ 2 + (y + 2)^2 = (x + 3)^2 + (y - 2)^2
which simplifies to -6x + 4y = 6x - 4y
ie y = 3x / 2 _______ (3)

You can substitute (3) into (1) or (2) to get a quadratic equation in x, and solve it to show that the only solution satisfying x > 0 is x = 2 * sqrt(3).

Is this clear enough?

 

[hatty]

yes it is

thanks alot buddy

 

[CM_Tutor]

No Problem.