MHB Find Intersection of Complex Number Loci Given w

AI Thread Summary
The discussion focuses on finding the intersection of two loci defined by the conditions \( |z-w|=|z-iw| \) and \( arg(z-w)=arg(iw) \) for a fixed complex number \( w \) in the first quadrant. Participants suggest that the intersection can be found geometrically by identifying the perpendicular bisector of the segment connecting points \( w \) and \( iw \), which represents points equidistant from both. Alternatively, an algebraic approach involves substituting \( w = a + ib \) and \( z = x + iy \) into the locus equations to derive Cartesian forms and solve them simultaneously. The original poster emphasizes that their challenge lies specifically in determining the intersection point rather than drawing the loci. Ultimately, both geometric and algebraic methods can yield the desired intersection in terms of \( w \).
Punch
Messages
44
Reaction score
0
w is a fixed complex number and \( 0<arg(w)<\frac{\pi}{2} \). Mark A and B, the points representing w and iw, on the Argand dagram. P represents the variable complex number z. Sketch on the same diagram, the locus of P in each of the following cases: (i) \( |z-w|=|z-iw| \) (ii) \(arg(z-w)=arg(iw)\)

Find in terms of w, the complex number representing the intersection of the two loci.

I have drawn the 2 locus already. But I do not know how to find the complex number representing the intersection of the 2 loci.
Do I form the equation of the 2 loci and then find the intersection by substituting one into the other?
 
Last edited:
Mathematics news on Phys.org
Use \ ( and \ ) without spaces to make your LaTeX work. As for the problem, remember that when you multiply complex numbers you rotate and expand/contract them, i.e., if \( z_1 = r_1 e^{ix_1} \text{ and } z_2 = r_2e^{i x_2} \text{ then } z_1z_2 = r_1r_2e^{i(x_1+x_2)} \). When you have \( |z-w| \) what you are measuring is the distance between \( z \text{ and } w \). Imposing that \( |z-w| = |z-iw| \) you want the locus of the points that are equally distant from \( w \text{ and } iw \).

Try working the second the same way. Remember the argument is the angle the complex number makes with the real axis.
 
Fantini said:
Use \ ( and \ ) without spaces to make your LaTeX work. As for the problem, remember that when you multiply complex numbers you rotate and expand/contract them, i.e., if \( z_1 = r_1 e^{ix_1} \text{ and } z_2 = r_2e^{i x_2} \text{ then } z_1z_2 = r_1r_2e^{i(x_1+x_2)} \). When you have \( |z-w| \) what you are measuring is the distance between \( z \text{ and } w \). Imposing that \( |z-w| = |z-iw| \) you want the locus of the points that are equally distant from \( w \text{ and } iw \).

Try working the second the same way. Remember the argument is the angle the complex number makes with the real axis.

Yup, I think you haven't read the next part I wrote. I completed drawing the locus and am facing difficulties solving the part which asks for a complex number representing the intersection of these 2 loci. "I have drawn the 2 locus already. But I do not know how to find the complex number representing the intersection of the 2 loci.
Do I form the equation of the 2 loci and then find the intersection by substituting one into the other?"
 
Geometrically, it will be the perpendicular passing through the midpoint connecting those two. Every point of it is equally distant to both. Algebraically, when you solve \( |z-w| = |z-iw| \) you should get two points, get the line passing through them and that's you answer. Since he asks for a sketch only, the geometric description should be easier to follow.
 
Fantini said:
Geometrically, it will be the perpendicular passing through the midpoint connecting those two. Every point of it is equally distant to both. Algebraically, when you solve \( |z-w| = |z-iw| \) you should get two points, get the line passing through them and that's you answer. Since he asks for a sketch only, the geometric description should be easier to follow.
The OP has said a few times now that s/he is NOT having trouble getting each locus, the trouble is getting the intersection of the two loci.

@OP: I have not looked closely, but you might be able to construct the intersection point geometrically in terms of w by using the isosceles triangles and symmetry that is present. Alternatively, an algebraic solution could be hammered out by substituting w = a + ib and z = x + iy into each locus to get the Cartesian equation and then solve using simultaneous equations and then link the answer back to w.
 
I'm sorry for not understanding the question properly, when I gave it further analysis I realized I was of no help.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top