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

[timon]

Homework Statement

Sketch the set of complex numbers z for which the following is true:

arg[(z+i)/(z-1)] = $$\pi$$/2

Homework 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]

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.

 

[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)?

 

[timon]

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.

 

[hunt_mat]

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.

 

[timon]

thats what i did! but the 'simple' part is eluding me. should i do arctan(imaginarypart/realpart)=pi/2 ?

 

[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?

 

[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}}]

 

[hunt_mat]

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?

 

[timon]

Solve for z(r):

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

I don't see how you got that. I've 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. :(

 

[hunt_mat]

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?

 

[timon]

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...?

 

[timon]

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?

 

[hunt_mat]

Sorry, I was in error, I corrected my post.

 

[jackmell]

I don't see how you got that. I've 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]

 

[hunt_mat]

That wrong, it should be
$$(x-1/2)^2+(y+1/2)^2=\frac{1}{2}$$

 

[timon]

i don't think i am allowed to use Mathematica on my exam ;)

 

[hunt_mat]

If you can understand my method then you should be sorted.

 

[Masja]

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..

 

[Masja]

the first one says: a(a-1)-b(b+1)=0
and #12 says: (b+1)(a-1)-ab=0

 

[jackmell]

i don't think i am allowed to use Mathematica on my exam ;)

For:

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

Note that:

$$x=-ty$$

or t=-x/y

Ok, now just substitute that into the expression for x.

 

[timon]

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?

ehm...no :(
i get r = (a-b)^(1/2)

 

[timon]

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..

he edited his post after i replied. my reply used the wrong expression. where do you study btw?

 

[hunt_mat]

You have:
$$a(a-1)+b(b+1)=a^{2}-a+b^{2}+b=\left( a-\frac{1}{2}\right)^{2}-\frac{1}{4}+\left( b+\frac{1}{2}\right)^{2}-\frac{1}{4}\Rightarrow \left( a-\frac{1}{2}\right)^{2}+\left( b+\frac{1}{2}\right)^{2}=\frac{1}{2}$$
This is the equation required, it is of a circle of radius $$1/\sqrt{2}$$ centred at the point (1/2,-1/2).
This is the

 

[Masja]

You have:
$$a(a-1)+b(b+1)=a^{2}-a+b^{2}+b=\left( a-\frac{1}{2}\right)^{2}-\frac{1}{4}+\left( b+\frac{1}{2}\right)^{2}-\frac{1}{4}\Rightarrow \left( a-\frac{1}{2}\right)^{2}+\left( b+\frac{1}{2}\right)^{2}=\frac{1}{2}$$
This is the equation required, it is of a circle of radius $$1/\sqrt{2}$$ centred at the point (1/2,-1/2).
This is the

but it is: a(a-1)-b(b+1)

 

[timon]

You have:
$$a(a-1)+b(b+1)=a^{2}-a+b^{2}+b=\left( a-\frac{1}{2}\right)^{2}-\frac{1}{4}+\left( b+\frac{1}{2}\right)^{2}-\frac{1}{4}\Rightarrow \left( a-\frac{1}{2}\right)^{2}+\left( b+\frac{1}{2}\right)^{2}=\frac{1}{2}$$
This is the equation required, it is of a circle of radius $$1/\sqrt{2}$$ centred at the point (1/2,-1/2).
This is the

aha! thanks a lot for the help, i doubt i would have figured this all out by myself :|

 

[timon]

BUT: then you have the same homework for calculus as me!

Indeed! Thats a funny coincidence. I am Dutch btw, are you not?

I didn't get the minus when i worked it out. I get:

$$a^2 - a + b^2 +b = 0$$

$$a(a-1) + b(b+1) = 0$$