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

1. Sep 25, 2010

### timon

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.

2. Sep 25, 2010

### 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)?

3. Sep 27, 2010

### 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.

4. Sep 27, 2010

### 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.

5. Sep 27, 2010

### timon

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

6. Sep 27, 2010

### 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?

7. Sep 27, 2010

### jackmell

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:
Code (Text):

z[r_] := (I*(1 + r))/(I*r - 1)
ParametricPlot[{Re[z[r]], Im[z[r]]},
{r, 0, 100}, PlotRange ->
{{-2, 2}, {-2, 2}}]

8. Sep 27, 2010

### 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?

9. Sep 29, 2010

### 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. :(

10. Sep 29, 2010

### 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?

11. Sep 29, 2010

### timon

I would say it means tan(pi/2) is equal to [imaginary part/real part] but tan(pi/2) is undefined...?

12. Sep 29, 2010

### 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?

13. Sep 29, 2010

### hunt_mat

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

14. Sep 29, 2010

### jackmell

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:

Code (Text):

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:

Code (Text):

ContourPlot[Arg[(x + I*y + I)/
(x + I*y - 1)] == Pi/2, {x, -2, 2},
{y, -2, 2}, Axes -> True]

Last edited: Sep 29, 2010
15. Sep 29, 2010

### hunt_mat

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

16. Sep 29, 2010

### timon

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

17. Sep 29, 2010

### hunt_mat

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

18. Sep 29, 2010

### Masja

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. Sep 29, 2010

### Masja

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

20. Sep 29, 2010

### jackmell

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.

Last edited: Sep 29, 2010