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Complex Numbers Easy Q watch

1. Graph the points z = x+iy that satisfy the condition |z + 2|= |z - 1| and interpret this geometrically.

How would I do this?

+REP!
2. Recall that . Substitute into the given equation, then look at the "new" values of x and y on each side. Then just square both sides and you have an equation in x and y which you can (hopefully!) draw in the plane.
3. Right so Z= x+iy;
|(x+2)+iy|=|(x-1)+iy|
You have to group them real and imaginary as above.
A modulus is the sqrt(real^2+imaginary^2),
sqrt(x^2+4x+4+y^2) = sqrt(x^2 -2x +1 +y^2)
Squaring again removes the square root sign.

From here you can simplify to find equation of the line,
x=--0.5
4. (Original post by impervious)
Right so Z= x+iy;
|(x+2)+iy|=|(x-1)+iy|
You have to group them real and imaginary as above.
A modulus is the sqrt(real^2+imaginary^2),
sqrt(x^2+4x+4+y^2) = sqrt(x^2 -2x +1 +y^2)
Squaring again removes the square root sign.

From here you can simplify to find equation of the line,
x=--0.5
I worked out x = -0.5, but then what about my iy ?
5. (Original post by Mos Def)
I worked out x = -0.5, but then what about my iy ?
Because the original equation given to you was |z+2|=|z-1|
You can see that it’s the loci between two real numbers, and as a result the line is vertical.
x=-.5 is the equation of the line, i.e a verticle line going through this point on the argand diagram.

Another way of looking at this is as follows.
The given equation can be written as:
|z-(-2)|=|z-(1)|
From which you can directly pull the point (-2,0) and (1,0).
The loci is the perpendicular bisector of these points.
6. (Original post by impervious)
Because the original equation given to you was |z+2|=|z-1|
You can see that it’s the loci between two real numbers, and as a result the line is vertical.
x=-.5 is the equation of the line, i.e a verticle line going through this point on the argand diagram.

Another way of looking at this is as follows.
The given equation can be written as:
|z-(-2)|=|z-(1)|
From which you can directly pull the point (-2,0) and (1,0).
The loci is the perpendicular bisector of these points.
Ok so i get that its a straight line at x=-0.5, but how did you know that the -2 in |z+2| is the x value for one point, and the same goes for 1.
7. (Original post by Mos Def)
Ok so i get that its a straight line at x=-0.5, but how did you know that the -2 in |z+2| is the x value for one point, and the same goes for 1.
Its sort of just how I was taught. Going about it the long was is fine, but if you can learn that method its quicker.

You know it is the x value as it was a real number if thats what you mean.

The notation say is to always have it |z-number|
So you get |z-(-2i)|=|z-(1)|
LHS the number is -2i, which by itself if the point (0,-2)
RHS is it number 1, which by itself is the point (1,0)
Therefore the loci will be the perpendicular bisector of these two co-ordinates, which will be the normal to the line through them, passing through the point midway between them.

Or you could solve it as follows
|z+2i|=|z-1|
|x+i(y+2)|=|(x-1)+iy|
sqrt[x^2 +y^2 +4y+4]= sqrt[x^2-2x+1+y^2]
x^2 +y^2 +4y+4= x^2-2x+1+y^2
4y+4=-2x+1
4y=-2x-3
8. (Original post by Mos Def)
Ok so i get that its a straight line at x=-0.5, but how did you know that the -2 in |z+2| is the x value for one point, and the same goes for 1.
Please tell me this isn't FP1!!! If it is, I am bloody screwed for my exam next week!
9. (Original post by Narik)
Please tell me this isn't FP1!!! If it is, I am bloody screwed for my exam next week!
Different boards have (although mostly the same) different content.
10. (Original post by impervious)
Its sort of just how I was taught. Going about it the long was is fine, but if you can learn that method its quicker.

You know it is the x value as it was a real number if thats what you mean.

The notation say is to always have it |z-number|
So you get |z-(-2i)|=|z-(1)|
LHS the number is -2i, which by itself if the point (0,-2)
RHS is it number 1, which by itself is the point (1,0)
Therefore the loci will be the perpendicular bisector of these two co-ordinates, which will be the normal to the line through them, passing through the point midway between them.

Or you could solve it as follows
|z+2i|=|z-1|
|x+i(y+2)|=|(x-1)+iy|
sqrt[x^2 +y^2 +4y+4]= sqrt[x^2-2x+1+y^2]
x^2 +y^2 +4y+4= x^2-2x+1+y^2
4y+4=-2x+1
4y=-2x-3
Oh right thanks +REP.

No its Calc II in uni maths, dno if its in your FP1 or not.

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