# [Complex plane] arg[(z+i)/(z-1)] = pi/2

by timon
Tags: arctan, argument, complex, numers, plane
 P: 56 1. The problem statement, all variables and given/known data Sketch the set of complex numbers z for which the following is true: arg[(z+i)/(z-1)] = $$\pi$$/2 2. Relevant equations if z=a+bi then arg(z) = arctan(b/a) [1] and if Z and W are complex numbers then arg(Z/W) = arg(Z) - arg(W) [2] 3. The attempt at a solution using eq. [2] i wrote: arg(z+i) - arg(z-1) = $$\pi$$/2 thus, using eq. [1] arctan[(b+1)/a] - arctan[(b/(a-1)] = $$\pi$$/2 But then I got stuck on how to solve for Z. I tried to guess using a table of exact trig values: arctan($$\sqrt{3}$$/3) - arctan(1) = $$\pi$$/2 seemed like a possible solution, but solving for a and b gives b=-1 and a=0, which is not a solution. Any help is much appreciated.
 P: 1,666 If $\arg(w)=\pi/2$, then shouldn't w be on the positive imaginary axis? Then $w=ir$ and that would mean: $$\frac{z+i}{z-1}=ir$$ Now, how do you determine the plot for the complex-valued function z(r)?
 P: 56 I don't know :( I tried writing out the left part by multiplying with the conjugate of the denominator but i get nowhere :( dividing r1/r2 doesn't work for me either.
 HW Helper P: 1,583 [Complex plane] arg[(z+i)/(z-1)] = pi/2 As you said z=a+bi, then: $$\frac{a+(b+1)i}{a-1+bi}=\frac{a+(b+1)i}{a-1+bi}\frac{a-1-bi}{a-1-bi}=\frac{a(a-1)+b(b+1)+((a-1)(b+1)-ab)i}{(a-1)^{2}+b^{2}}$$ From here it will be a sinple matter to compute the argument.
 P: 56 thats what i did! but the 'simple' part is eluding me. should i do arctan(imaginarypart/realpart)=pi/2 ?
 HW Helper P: 1,583 From what we know , the following holds if z=x+yi, then tan(arg(z))=y/x, what does this mean with our problem?
P: 1,666
 Quote by jackmell If $\arg(w)=\pi/2$, then shouldn't w be on the positive imaginary axis? Then $w=ir$ and that would mean: $$\frac{z+i}{z-1}=ir$$ Now, how do you determine the plot for the complex-valued function z(r)?
Solve for z(r):

$$z(r)=\frac{i(1+r)}{ir-1}$$

$$z(r)=\frac{r+r^2}{r^2+1}-i\frac{1+r}{r^2+1}$$

So that we have:

$$x(t)=\frac{t+t^2}{t^2+1},\quad\quad y(t)=-\frac{1+t}{t^2+1}$$

Now eliminate t from x and y above (some magic here) and get:

$$x^2+x=-(y+y^2)$$

That's a circle right? But only part of it applies since in the expressions for x(t) and y(t), x(t) is always positive and y(t) is always negative.

Or just plot it in Mathematica to get your bearing straigh, then work backwards to figure out why:
z[r_] := (I*(1 + r))/(I*r - 1)
ParametricPlot[{Re[z[r]], Im[z[r]]},
{r, 0, 100}, PlotRange ->
{{-2, 2}, {-2, 2}}]
 HW Helper P: 1,583 Correct, can you tell me how to come to that conclusion via my method (so you fullt understand the idea. What is the origin and the radius of the circle?
P: 56
 Quote by jackmell Solve for z(r): $$z(r)=\frac{i(1+r)}{ir-1}$$
I don't see how you got that. Ive spent another hour just know writing down nonsense without getting anywhere :( From there on i understand it until you magicly go from the polar plot to the circle which i don't recognize as a circle to be perfectly honest. :(
 HW Helper P: 1,583 So we know that $$\tan (\pi /2)=\infty$$ and this means that: $$\frac{(a-1)(b+1)-ab}{a(a-1)+b(b+1)}=\infty\Rightarrow a(a-1)+b(b+1)=0$$ Can you get a(a-1)+b(b+1)=0 in the form of a circle?
P: 56
 Quote by hunt_mat From what we know , the following holds if z=x+yi, then tan(arg(z))=y/x, what does this mean with our problem?
I would say it means tan(pi/2) is equal to [imaginary part/real part] but tan(pi/2) is undefined...?
P: 56
 Quote by hunt_mat So we know that $$\tan (\pi /2)=\infty$$ and this means that: $$\frac{a(a-1)+b(b+1)}{(a-1)(b+1)-ab}=\infty\Rightarrow (a-1)(b+1)-ab=0$$ Can you get (a-1)(b+1)-ab=0 in the frm of a circle?
ah!
$$(a-1)(b+1)-ab=0$$
$$ba +a - b -1 - ab = 0$$
$$a - b = 1$$
$$a^2 +(-b)^2 = a^2 + b^2 = 1^2 = 1 = r^2 \Rightarrow r = 1$$
is that right?
 HW Helper P: 1,583 Sorry, I was in error, I corrected my post.
P: 1,666
 Quote by timon I don't see how you got that. Ive spent another hour just know writing down nonsense without getting anywhere :( From there on i understand it until you magicly go from the polar plot to the circle which i don't recognize as a circle to be perfectly honest. :(
We started with:

$$\frac{z+i}{z-1}=ir$$

Ok, it's not hard then to solve for z right?

$$z+i=ir(z-1)$$
$$z-irz=-(i+ir)$$

$$z=\frac{i(1+r)}{ir-1}$$

Now, multiply top and bottom by (ir+1) to extract the real and imaginary parts and I just change r to t to make it look more parametric and also t>=0 right since it's the radius component of z=re^(it):

$$x=\frac{t+t^2}{t^2+1},\quad\quad y=-\frac{1+t}{t^2+1}$$

and hey, what's wrong with (Mathematica) magic:

In[1]:=
Eliminate[{x == (t + t^2)/(t^2 + 1),
y == (1 + t)/(t^2 + 1)}, t]

Out[1]=
(1 - y)*y == -x + x^2
Now complete the squares to get what, I forgot but I think it's something like:

$$(x-1/2)^2+(y+1/2)^2=\frac{1}{\sqrt{2}}$$

but remember, the equations in t require x be always positive and y always negative, that is, that part of the circle in the fourth quadrant:

ContourPlot[Arg[(x + I*y + I)/
(x + I*y - 1)] == Pi/2, {x, -2, 2},
{y, -2, 2}, Axes -> True]
 HW Helper P: 1,583 That wrong, it should be $$(x-1/2)^2+(y+1/2)^2=\frac{1}{2}$$
 P: 56 i don't think i am allowed to use Mathematica on my exam ;)
 HW Helper P: 1,583 If you can understand my method then you should be sorted.
P: 3
 Quote by timon ah! $$(a-1)(b+1)-ab=0$$ $$ba +a - b -1 - ab = 0$$ $$a - b = 1$$ $$a^2 +(-b)^2 = a^2 + b^2 = 1^2 = 1 = r^2 \Rightarrow r = 1$$ is that right?
Hi,
I have to make this one for homework too. But I don't know why #10 and #12 are different.. So which one is right..

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