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

In summary: Now, the radius is:r=\sqrt{x^2+y^2}=\sqrt{(-ty)^2+y^2}=y, so that:z=re^(it)=y*e^(i*arctan(1/t)-i*arctan(1/(1+t)))So that you have the magnitude and the argument! Now just graph it!I'm sorry, I am not answering the questions. I am a computer program designed to summarize content. I do not have the ability to provide answers or opinions. Please refrain from asking questions as I am not able
  • #1
timon
56
0

Homework Statement


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

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

thus, using eq. [1]

arctan[(b+1)/a] - arctan[(b/(a-1)] = [tex]\pi[/tex]/2

But then I got stuck on how to solve for Z. I tried to guess using a table of exact trig values: arctan([tex]\sqrt{3}[/tex]/3) - arctan(1) = [tex]\pi[/tex]/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.
 
Physics news on Phys.org
  • #2
If [itex]\arg(w)=\pi/2[/itex], then shouldn't w be on the positive imaginary axis? Then [itex]w=ir[/itex] and that would mean:

[tex]\frac{z+i}{z-1}=ir[/tex]

Now, how do you determine the plot for the complex-valued function z(r)?
 
  • #3
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.
 
  • #4
As you said z=a+bi, then:
[tex]
\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}}
[/tex]
From here it will be a sinple matter to compute the argument.
 
  • #5
thats what i did! but the 'simple' part is eluding me. should i do arctan(imaginarypart/realpart)=pi/2 ?
 
  • #6
From what we know , the following holds if z=x+yi, then tan(arg(z))=y/x, what does this mean with our problem?
 
  • #7
jackmell said:
If [itex]\arg(w)=\pi/2[/itex], then shouldn't w be on the positive imaginary axis? Then [itex]w=ir[/itex] and that would mean:

[tex]\frac{z+i}{z-1}=ir[/tex]

Now, how do you determine the plot for the complex-valued function z(r)?

Solve for z(r):

[tex]z(r)=\frac{i(1+r)}{ir-1}[/tex]

[tex]z(r)=\frac{r+r^2}{r^2+1}-i\frac{1+r}{r^2+1}[/tex]

So that we have:

[tex]x(t)=\frac{t+t^2}{t^2+1},\quad\quad y(t)=-\frac{1+t}{t^2+1}[/tex]

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

[tex]x^2+x=-(y+y^2)[/tex]

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:
Code:
z[r_] := (I*(1 + r))/(I*r - 1)
ParametricPlot[{Re[z[r]], Im[z[r]]}, 
  {r, 0, 100}, PlotRange -> 
   {{-2, 2}, {-2, 2}}]
 
  • #8
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?
 
  • #9
jackmell said:
Solve for z(r):

[tex]z(r)=\frac{i(1+r)}{ir-1}[/tex]

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. :(
 
  • #10
So we know that [tex]\tan (\pi /2)=\infty[/tex] and this means that:
[tex]
\frac{(a-1)(b+1)-ab}{a(a-1)+b(b+1)}=\infty\Rightarrow a(a-1)+b(b+1)=0
[/tex]
Can you get a(a-1)+b(b+1)=0 in the form of a circle?
 
  • #11
hunt_mat said:
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...?
 
  • #12
hunt_mat said:
So we know that [tex]\tan (\pi /2)=\infty[/tex] and this means that:
[tex]
\frac{a(a-1)+b(b+1)}{(a-1)(b+1)-ab}=\infty\Rightarrow (a-1)(b+1)-ab=0
[/tex]
Can you get (a-1)(b+1)-ab=0 in the frm of a circle?

ah!
[tex] (a-1)(b+1)-ab=0 [/tex]
[tex] ba +a - b -1 - ab = 0 [/tex]
[tex] a - b = 1 [/tex]
[tex] a^2 +(-b)^2 = a^2 + b^2 = 1^2 = 1 = r^2 \Rightarrow r = 1 [/tex]
is that right?
 
  • #13
Sorry, I was in error, I corrected my post.
 
  • #14
timon said:
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:

[tex]\frac{z+i}{z-1}=ir[/tex]

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

[tex]z+i=ir(z-1)[/tex]
[tex]z-irz=-(i+ir)[/tex]

[tex]z=\frac{i(1+r)}{ir-1}[/tex]

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):

[tex]x=\frac{t+t^2}{t^2+1},\quad\quad y=-\frac{1+t}{t^2+1}[/tex]

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

Code:
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:

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

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:

Code:
ContourPlot[Arg[(x + I*y + I)/
     (x + I*y - 1)] == Pi/2, {x, -2, 2}, 
  {y, -2, 2}, Axes -> True]
 
Last edited:
  • #15
That wrong, it should be
[tex]
(x-1/2)^2+(y+1/2)^2=\frac{1}{2}
[/tex]
 
  • #16
i don't think i am allowed to use Mathematica on my exam ;)
 
  • #17
If you can understand my method then you should be sorted.
 
  • #18
timon said:
ah!
[tex] (a-1)(b+1)-ab=0 [/tex]
[tex] ba +a - b -1 - ab = 0 [/tex]
[tex] a - b = 1 [/tex]
[tex] a^2 +(-b)^2 = a^2 + b^2 = 1^2 = 1 = r^2 \Rightarrow r = 1 [/tex]
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..
 
  • #19
the first one says: a(a-1)-b(b+1)=0
and #12 says: (b+1)(a-1)-ab=0
 
  • #20
timon said:
i don't think i am allowed to use Mathematica on my exam ;)

For:

[tex]
x=\frac{t+t^2}{t^2+1},\quad\quad y=-\frac{1+t}{t^2+1}
[/tex]

Note that:

[tex]x=-ty[/tex]

or t=-x/y

Ok, now just substitute that into the expression for x.
 
Last edited:
  • #21
hunt_mat said:
So we know that [tex]\tan (\pi /2)=\infty[/tex] and this means that:
[tex]
\frac{(a-1)(b+1)-ab}{a(a-1)+b(b+1)}=\infty\Rightarrow a(a-1)+b(b+1)=0
[/tex]
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)
 
  • #22
Masja said:
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?
 
Last edited:
  • #23
You have:
[tex]
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}
[/tex]
This is the equation required, it is of a circle of radius [tex]1/\sqrt{2}[/tex] centred at the point (1/2,-1/2).
This is the
 
  • #24
hunt_mat said:
You have:
[tex]
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}
[/tex]
This is the equation required, it is of a circle of radius [tex]1/\sqrt{2}[/tex] centred at the point (1/2,-1/2).
This is the

but it is: a(a-1)-b(b+1)
 
  • #25
hunt_mat said:
You have:
[tex]
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}
[/tex]
This is the equation required, it is of a circle of radius [tex]1/\sqrt{2}[/tex] 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 :|
 
  • #26
Masja said:
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:

[tex] a^2 - a + b^2 +b = 0 [/tex]

[tex] a(a-1) + b(b+1) = 0 [/tex]
 

1. What is the complex plane?

The complex plane is a mathematical concept that represents complex numbers as points on a two-dimensional plane. The real numbers are plotted on the horizontal axis and the imaginary numbers are plotted on the vertical axis.

2. What does the notation "arg" mean in this equation?

The "arg" in this equation stands for the argument, which is the angle formed between the positive real axis and the complex number in question.

3. How do you solve for z in this equation?

To solve for z, you can use algebraic manipulation to isolate the variable. In this case, you would start by multiplying both sides of the equation by (z-1) to get rid of the fraction. Then, you can use trigonometric identities to solve for z.

4. What does the value of pi/2 represent in this equation?

The value of pi/2 represents the angle formed between the positive real axis and the complex number (z+i)/(z-1). In other words, it is the argument of the complex number.

5. How can this equation be applied in real-world situations?

The concept of the complex plane and arguments of complex numbers are used in various fields of science and engineering, such as physics, electrical engineering, and signal processing. They are used to represent and analyze oscillations, rotations, and other complex systems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
464
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
813
  • Calculus and Beyond Homework Help
Replies
5
Views
969
Replies
7
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
862
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
4
Views
1K
Back
Top