how to solve this complex number question..

transgalactic · #1 $Old$ 11-20-2008, 11:08 AM

http://img353.imageshack.us/img353/672/85253506or3.gif

in normal equation i equalize the "Real" part with the real part
and the "Im" part with the Im part on the other size of the equation
but here there is | | part

which makes every thing a^2 + b^2 and it turns everything to "real"

??

Opalg · #2 $Old$ 11-20-2008, 02:15 PM

$|z+i| = |z-1| = \bigl|\tfrac1{\sqrt2}+\tfrac1{\sqrt2}i\bigr|$

This is a problem that is best done geometrically. Remember that |z-a| is the distance from z to a. So $|z+i| = |z-1|$ means that the distance from z to -i is the same as the distance from z to 1. Therefore z lies on the perpendicular bisector of the line joining -i to 1. The other part of the equation means that z is also on the perpendicular bisector of the line joining 1 to $\tfrac1{\sqrt2}+\tfrac1{\sqrt2}i$ , and that pins down the location of z uniquely.

On the other hand, if you want to solve the equation algebraically, write z=x+iy. Then the equation $|z+i| = |z-1|$ becomes $|x+i(y+1)| = |(x-1)+iy|$ . Square both sides, and that tells you that $x^2+(y+1)^2 = (x-1)^2+y^2$ . Multiply out the brackets and you'll find that that equation simplifies to the equation of a straight line (which is in fact the perpendicular bisector of the line joining -i to 1, as in the geometric solution).

transgalactic · #3 $Old$ 11-20-2008, 02:26 PM

how do i solve such a thing

http://img353.imageshack.us/img353/672/85253506or3.gif[/quote]
Any number of ways. First, note that you really have two equations there. There may not exist z satisfying both of them.

For one of them, say |z+ i|= |z- 1|, let z= x+ iy and expand it: z+ i= x+ (y+1)i and |z+ i|= [itex]\sqrt{x^2+ (y+1)^2}[/itex] while z-1= (x- 1)+ iy so |z-1|= [itex]\sqrt{(x-1)^2+ y^}[/itex] |z+ i|= |z- 1|, then, is the same as [itex]\sqrt{x^2+ (y+1)^2}= \sqrt{(x-1)^2+ y^2}[/itex] so [itex]x^2+ (y+1)^2= (x-1)^2+ y^2[/itex]. If you multiply those out the squares cancel and you should see that is a straight line in the xy-plane. You could do the same thing with [itex]|z+1|= |1/\sqrt{2}+ i/\sqrt{2}- z| and again see the it is a straight line in the xy-plane. The solution of the entire system is the point where those two lines intersect.

Of you could do it geometrically: |a- b| can be interpreted as the distance between points a and b in the complex plane. In particular, |z-1|= |z+ i| is the set of points that are equidistant from -1 and i or, in terms of (x,y), (-1, 0) and (0, 1). That is, of course, the perpendicular bisector on that line segement. Similarly, [itex]|z-1|= |1/\sqrt{2}+ i/\sqrt{2}- z| is the set of points equidistant from 1 and [itex]1/\sqrt{2}+ i/\sqrt{2}[/itex]: the perpendicular bisector of the line segment from (1, 0) to [itex](1/\sqrt{2}, 1/\sqrt{2})[/itex]. The point satisfying the original system is the point where those two lines intersect.

Mathstud28 · #4 $Old$ 11-20-2008, 02:47 PM

Quote:

Originally Posted by transgalactic $View Post$

which makes every thing a^2 + b^2 and it turns everything to "real"

??

Be careful, in all likelihood you just made a typo, but the norm of a n-tuple $\bold{x}=\left(a_1,a_2,\cdots,a_n\right)$ is given by $, \left|\bold{x}\right|=\sqrt{\sum_{i=1}^nx_i^2}$ , so since the complex numbers are an ordered pair (a 2-tuple) we have that $|z|={\color{red}\sqrt{a^2+b^2}}$

Quote:

Originally Posted by transgalactic $View Post$

how do i solve such a thing

http://img353.imageshack.us/img353/672/85253506or3.gif
Any number of ways. First, note that you really have two equations there. There may not exist z satisfying both of them.

For one of them, say |z+ i|= |z- 1|, let z= x+ iy and expand it: z+ i= x+ (y+1)i and |z+ i|= [itex]\sqrt{x^2+ (y+1)^2}[/itex] while z-1= (x- 1)+ iy so |z-1|= [itex]\sqrt{(x-1)^2+ y^}[/itex] |z+ i|= |z- 1|, then, is the same as [itex]\sqrt{x^2+ (y+1)^2}= \sqrt{(x-1)^2+ y^2}[/itex] so [itex]x^2+ (y+1)^2= (x-1)^2+ y^2[/itex]. If you multiply those out the squares cancel and you should see that is a straight line in the xy-plane. You could do the same thing with [itex]|z+1|= |1/\sqrt{2}+ i/\sqrt{2}- z| and again see the it is a straight line in the xy-plane. The solution of the entire system is the point where those two lines intersect.

Of you could do it geometrically: |a- b| can be interpreted as the distance between points a and b in the complex plane. In particular, |z-1|= |z+ i| is the set of points that are equidistant from -1 and i or, in terms of (x,y), (-1, 0) and (0, 1). That is, of course, the perpendicular bisector on that line segement. Similarly, [itex]|z-1|= |1/\sqrt{2}+ i/\sqrt{2}- z| is the set of points equidistant from 1 and [itex]1/\sqrt{2}+ i/\sqrt{2}[/itex]: the perpendicular bisector of the line segment from (1, 0) to [itex](1/\sqrt{2}, 1/\sqrt{2})[/itex]. The point satisfying the original system is the point where those two lines intersect.

And on this site the latex tags are [math] [/math not [itex][/itex]