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[Complex plane] arg[(z+i)/(z1)] = pi/2 
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#1
Sep2510, 11:13 AM

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)/(z1)] = [tex]\pi[/tex]/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(z1) = [tex]\pi[/tex]/2 thus, using eq. [1] arctan[(b+1)/a]  arctan[(b/(a1)] = [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. 


#2
Sep2510, 12:26 PM

P: 1,666

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}{z1}=ir[/tex] Now, how do you determine the plot for the complexvalued function z(r)? 


#3
Sep2710, 01:05 PM

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.



#4
Sep2710, 01:36 PM

HW Helper
P: 1,583

[Complex plane] arg[(z+i)/(z1)] = pi/2
As you said z=a+bi, then:
[tex] \frac{a+(b+1)i}{a1+bi}=\frac{a+(b+1)i}{a1+bi}\frac{a1bi}{a1bi}=\frac{a(a1)+b(b+1)+((a1)(b+1)ab)i}{(a1)^{2}+b^{2}} [/tex] From here it will be a sinple matter to compute the argument. 


#5
Sep2710, 01:48 PM

P: 56

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



#6
Sep2710, 01:58 PM

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?



#7
Sep2710, 04:50 PM

P: 1,666

[tex]z(r)=\frac{i(1+r)}{ir1}[/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:



#8
Sep2710, 04:52 PM

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?



#9
Sep2910, 10:04 AM

P: 56




#10
Sep2910, 10:08 AM

HW Helper
P: 1,583

So we know that [tex]\tan (\pi /2)=\infty[/tex] and this means that:
[tex] \frac{(a1)(b+1)ab}{a(a1)+b(b+1)}=\infty\Rightarrow a(a1)+b(b+1)=0 [/tex] Can you get a(a1)+b(b+1)=0 in the form of a circle? 


#11
Sep2910, 10:09 AM

P: 56




#12
Sep2910, 10:13 AM

P: 56

[tex] (a1)(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
Sep2910, 10:15 AM

HW Helper
P: 1,583

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



#14
Sep2910, 10:19 AM

P: 1,666

[tex]\frac{z+i}{z1}=ir[/tex] Ok, it's not hard then to solve for z right? [tex]z+i=ir(z1)[/tex] [tex]zirz=(i+ir)[/tex] [tex]z=\frac{i(1+r)}{ir1}[/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:
[tex](x1/2)^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:



#15
Sep2910, 10:22 AM

HW Helper
P: 1,583

That wrong, it should be
[tex] (x1/2)^2+(y+1/2)^2=\frac{1}{2} [/tex] 


#16
Sep2910, 11:28 AM

P: 56

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



#17
Sep2910, 12:03 PM

HW Helper
P: 1,583

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



#18
Sep2910, 12:06 PM

P: 3

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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