Solve this Complex Equation

[@txikitofandango]

Oh nice. I obtained the circle in basically the opposite way, by noting that, if z = a + bi, and the angle between z-1 and z+1 is a right angle, then the dot product (a+1,b) . (a-1,b) = 0 implies that a^2 + b^2 = 1.

[@DrBarker]

This is very neat, a nice use of the dot product!

[@mrigayu]

A brilliant application of complex numbers and geometry! I tend to approach these problems through algebraic methods first (as another user has suggested in the comments), but I really like the elegance of geometric and/or pictorial approaches like this as it provides a really strong intuition. Especially when you reach that classic circle theorem we all learn in school and also for the case where Im(z) < 0

[@沈博智-x5y]

From a purely algebraic view: arg(z-1) = arg(z+1) + pi/2 where z eq -1 , 1 arg(z-1) - arg(z+1) = pi/2 rearranging equation arg((z-1)/(z+1)) = pi/2 (difference of arguments is the argument of the quotient) Since z +1 eq 0 or z eq -1, this implies conj(z) eq -1 or conj(z) + 1 eq 1 Multiply the numerator and denominator by (conj(z) + 1) arg((z-1)(conj(z) + 1)/(z+1)(conj(z) + 1)) = pi/2 (z+1)(conj(z) + 1) = |z+1|^2 Pf: LHS = (z+1)(conj(z) + 1) = zconj(z) + z + conj(z) + 1 = (x+iy)(x-iy) + (x+iy) + (x - iy) + 1 where z = x + iy where x and y are real = x^2 + y^2 + 2x + 1 = x^2 + 2x + 1 + y^2 = (x+1)^2 + y^2 = |z+1|^2 so this implies: arg((|z|^2 + z - conj(z) - 1)/|z+1|^2) = pi/2 arg(|z|^2 - 1 + 2Im(z)) = pi/2 since z+1 eq 0 => |z+1|^2 > 0 This only happens when |z|^2 - 1 = 0 and 2Im(z) > 0 Or when |z|^2 = 1 and Im(z) > 0 Or when |z| = 1 and Im(z) > 0 since |z| >= 0 In other words, the upper half of a semi-circle, radius 1, centred at the origin, with poles at z = -1, 1

[@DrBarker]

This is an interesting alternative approach. I thought about doing this algebraically using arg((z-1)/(z+1)) = pi/2 so (z-1)/(z+1) is pure imaginary. Then we can write z = a + bi, and in order for (z-1)/(z+1) to be pure imaginary, we can show that a^2 + b^2 = 1. There are some technicalities though with the arg(z-1) - arg(z+1) = arg((z-1)/(z+1)) step. This isn't always true, depending on which quadrant z lies in.

[@沈博智-x5y]

That's a good point with the technicality there. Would theta = arg(z) - arg(w) => theta = arg(z/w) work provided that arg(z) - arg(w) \in (-pi, pi] and the original equation when rearranged has arg(z-1) - arg(z+1) = pi/2 \in (-pi, pi] would that be enough to deduce that the argument identity can be used?

[@sternmg]

I used this alternative geometrical approach: Start with an arbitrary z and consider all directions (oriented rays) emanating from z+1. For each direction, construct the corresponding direction at z-1 with its arg increased by π/2, per the given. For z to be a solution, the origin must be at the intersection of the _reverse_ of both rays. This is only satisfiable for 0 < arg(z) < π. We obtain the condition |z| ≡ 1 from Thales’s theorem, just like in the video. No other solutions exist: For z∈ℝ, we get the degenerate cases where arg isn’t well-defined for z+1 or z-1, and for all other z, the intersection would occur in the _forward_ direction of one or both rays, meaning the intersection cannot be the origin that determined the args of z±1.

[@bjornfeuerbacher5514]

Yet another possible approach: First we assume that Im(z) > 0. The other cases have to be dealt with separately, just as in the video. We write z = a + ib and thus have arg(z-1) = arctan(y/(x-1)) and arg(z+1) = arctan(y/(x+1)), where we choose the arctan so that it gives an angle between pi/2 and pi for negative arguments. Inserting this into the given equation: arctan(b/(a-1)) = arctan(b/(a+1)) + pi/2. Now take the tan on both sides and use that tan(alpha + pi/2) = -1/tan(alpha). So we get: b/(a-1) = -(a+1)/b. That equation can easily be simplified to a² + b² = 1, i. e. the equation of a circle, and due to the condition Im(z) > 0, we again get the upper half circle.

[@mcqueen424]

Dr Barker what is your main field? I’m a grad student in numerical PDEs right now

[@kewalmer7225]

this question is a no-brainer, the angle in a circle is pi/2 thus the locus is just a semi-circle with radius 1unit . If someone isnt able to visualize it right away then we can also solve it purely with algebra

[@TaladrisKpop]

arg(z)-arg(w)=arg(z/w) (mod 2pi) so arg((z-1)/(z+1))=pi/2, so you can write (z-1)/(z+1) = r exp(ipi/2)=ri, and then solve for z. The geometric method is nice too.

[@RGP_Maths]

These Argand locus problems were quite common in the A Level Further Pure Mathematics text books, so I solved it in my head from the thumbnail, by Dr Barker's method.

[@digxx]

Pi/2=arg(z-1)-arg(z+1)=arg((z-1)/(z+1)). Writing z=x+iy we have (z-1)/(z+1)=(x^2-1+y^2)/((x+1)^2+y^2) + i 2y/((x+1)^2+y^2) after some calculation. If the argument is Pi/2, this means that this number is purely imaginary with positive imaginary part i.e. y>0. Solving x^2-1+y^2=0 gives y=sqrt(1-x^2).

[@Happy_Abe]

Why at 8:26 is Z-1 on the right and Z+1 on the left?

[@kipu44]

I was confused too. See 3:30. The vectors were shifted in opposite directions which resulted in swapping the angles.

[@Happy_Abe]

@@kipu44gotcha thank you!

[@evankalis]

I considered Z as a number of the form r*e^(i*theta) then just solved it in my head. Found one solution and ruled out the others. Ot ended up being purely imaginary and only in the top half plane. I then figured out the solution by scaling two unit vectors offset by pi/2 radians and boom the answer is i