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Locus of arg(z-i)=pi/2

Hello there.

I'm looking to find the locus of arg(z-i)=pi/2

I've been trying to convert this to something in terms of x and y to no avail, i.e.:

arg(x + iy -i) = pi/2
arg(x + (y - 1)i) = pi/2
arctan((y - 1)/x)=pi/2
(y - 1)/x = tan(pi/2)

But tan(pi/2) is undefined, so this must be the wrong approach.

Any help would be much appreciated! :smile:

Reply 1

Bugzy

10

Plot it - you will see it when you do.

Reply 2

|| Mod Zed ||

So pi/2 is going to be the angle from the positive x-axis going in an anti-clockwise direction.

The arg(z-i), just means you translate it one unit in the 'i' direction. So it will move one unit upwards and the angle parallel to the positive x-axis will be pi/2.

I can't really explain it anymore than that.

Reply 3

rnd

6

Harry1W

Hello there.

I'm looking to find the locus of arg(z-i)=pi/2

Can you draw arg(w)=pi/2 ?
If w=z-i then z=w+i so its just your last sketch moved up 1 in the i direction.

Reply 4

Tzar_Chasm

Please don't tell me this is C4. Or I may actually cry.

Reply 5

jammer

Please don't tell me this is FP1. Or I WILL purge all over my keyboard and possibly cry simultaneously.

Reply 6

Bugzy

10

It is fp2 new spec (fp3 old spec?)

edit - when has complex numbers ever been in c4?

Reply 7

generalebriety

16

Tzar_Chasm

Please don't tell me this is C4. Or I may actually cry.

Why would you think it was C4? At least the person below you who thinks it's FP1 has seen complex numbers before. This is FP2/3 (depending on syllabus).

Harry1W

Hello there.

I'm looking to find the locus of arg(z-i)=pi/2

I've been trying to convert this to something in terms of x and y to no avail, i.e.:

arg(x + iy -i) = pi/2
arg(x + (y - 1)i) = pi/2
arctan((y - 1)/x)=pi/2
(y - 1)/x = tan(pi/2)

But tan(pi/2) is undefined, so this must be the wrong approach.

Any help would be much appreciated! :smile:

In a way, that's fine. cot(pi/2) is perfectly well defined and is 0, so x/(y-1) = cot(pi/2) = 0, and so x = 0, giving us the imaginary axis. Unfortunately, this gives us too much - try plugging in the point z = 0 (which is definitely on the imaginary axis) and you'll see why. Algebraic methods of working out loci always have this problem.

Remember back to graph transformations? You used to play about with things of the form y = f(x) and shift them to y = f(x-a). Well, do the same thing here. Let arg(z) = pi/2 - which complex numbers satisfy this property? Now shift to arg(z-i) = pi/2. You should see part of the imaginary axis coming out of this (as we worked out above), but not all of it.

Reply 8

Tzar_Chasm

generalebriety

Why would you think it was C4? At least the person below you who thinks it's FP1 has seen complex numbers before.

Because I have a desperately large amount of C4 learning to by Monday. Seriously, I have no idea what is in C4 except there is a hell of a lot of integration.

Reply 9

Harry1W

OP

4

OK. So essentially you're arguing:

Let: f(z) = arg(z)
Then: arg(z - i) = f(z - i)

...and then the usual rules of transformation apply to give a half line x=0, y>1

The problem (I tentatively suggest) I have with this is that: z = f( x, y) so I'm not sure that the simple rules of transformation apply. When I graph the function in Wolfram Alpha, you can some kind of transformation of a tan curve, not this half line: http://www10.wolframalpha.com/input/?i=arg[z-i]%3Dpi%2F2

P.S.: Don't panic! This question is taken from a first year Oxford lecture question sheet. I've done Maths & Further Maths A-level and don't remember any question as tricksy as this one in the complex no's section!

Reply 10

The Financial Astrologer

Harry1W

Hello there.

I'm looking to find the locus of arg(z-i)=pi/2

I've been trying to convert this to something in terms of x and y to no avail, i.e.:

arg(x + iy -i) = pi/2
arg(x + (y - 1)i) = pi/2
arctan((y - 1)/x)=pi/2
(y - 1)/x = tan(pi/2)

But tan(pi/2) is undefined, so this must be the wrong approach.

Any help would be much appreciated! :smile:

That is correct mate.

(y-1)/x = undefined

Therefore in order for (y-1)/x to be undefined x=0

Since (y-1)/0=undefined

Hence half line starts at (0,1) downwards.

Reply 11

Trevor 12345

3

Harry1W

OK. So essentially you're arguing:

Let: f(z) = arg(z)
Then: arg(z - i) = f(z - i)

...and then the usual rules of transformation apply to give a half line x=0, y>1

The problem (I tentatively suggest) I have with this is that: z = f( x, y) so I'm not sure that the simple rules of transformation apply. When I graph the function in Wolfram Alpha, you can some kind of transformation of a tan curve, not this half line: http://www10.wolframalpha.com/input/?i=arg[z-i]%3Dpi%2F2

P.S.: Don't panic! This question is taken from a first year Oxford lecture question sheet. I've done Maths & Further Maths A-level and don't remember any question as tricksy as this one in the complex no's section!

hmmm I would say that this question is significantly easier than the standard edexcel FM questions on said subject.

Reply 12

Trevor 12345

3

I would answer the questions by considering the the problem as one of vectors... i.e. you want the set of points such that the vector drawn from 0+i to z is at an angle of pi/2 with a line drawn through 0 + i parallel to the real axis.

Reply 13

Harry1W

OP

4

The Financial Astrologer

That is correct mate.

(y-1)/x = undefined

Therefore in order for (y-1)/x to be undefined x=0

Since (y-1)/0=undefined

Hence half line starts at (0,1) downwards.

I like this explanation. Could you just explain how it is apparent from this explanation that y>1 (i.e. the half-line starts at (0,1))?

Also, if someone could explain why Wolfram Alpha gave the crazy tan family curve (http://www10.wolframalpha.com/input/?i=arg[z-i]%3Dpi%2F2), I'll sleep very soundly. :yes:

Mrm.

hmmm I would say that this question is significantly easier than the standard edexcel FM questions on said subject.

Sorry! It's a few years since I've done this, so perhaps I'm probably out of touch. Panic!

generalebriety

In a way, that's fine. cot(pi/2) is perfectly well defined and is 0, so x/(y-1) = cot(pi/2) = 0, and so x = 0, giving us the imaginary axis. Unfortunately, this gives us too much - try plugging in the point z = 0 (which is definitely on the imaginary axis) and you'll see why. Algebraic methods of working out loci always have this problem.

I can see that this works algebraically, but why is OK to essentially divide by an undefined quantity to get another that is defined (as zero)?

Reply 14

generalebriety

16

Harry1W

I can see that this works algebraically, but why is OK to essentially divide by an undefined quantity to get another that is defined (as zero)?

Define

\displaystyle \cot(\pi/2) = \lim_{x\to \pi/2} (\tan(x))^{-1}

. The function tan may be discontinuous at pi/2 (i.e. the limit of tan(x) does not exist as x --> pi/2, because from the left it's heading to +infinity and from the right it's heading to -infinity), but cot certainly isn't (because the reciprocal tends to 0 in both cases).

The problem here is the same as the problem of writing the equation of a vertical line in the form y = mx + c. You can't, but if you try and write it in the form x = ny + d, suddenly you can. Why?

Harry1W

The problem (I tentatively suggest) I have with this is that: z = f( x, y) so I'm not sure that the simple rules of transformation apply.

Well, at the very least, the transformation f(x, y) --> f(x, y-1) is sufficiently simple for this not to matter. If you're at university, I'm sure you can convince yourself of this!

(Edit: in fact, just put Z = z-i, and solve arg(Z) = pi/2 for Z. Then you get Z being anything on the positive imaginary axis, i.e. Z = iy for y > 0. But Z = z-i...

This is how all such transformations work. Suppose you have f(x) = a, where x is a vector of variables, e.g. (x, y), and we want to solve f(x - k) = a, where k is a constant vector. Simply define X = x - k, solve f(X) = a, and each solution X = X₀ to this equation gives a solution x = X₀ + k. In general, you can solve f(g(x)) = k by putting X = g(x) and, for each solution X = X₀, solve the equation g(x) = X₀. This normally turns out nicely if g is nice (and invertible).)

((Edit 2: I hate bolding.))

Reply 15

generalebriety

16

Harry1W

When I graph the function in Wolfram Alpha, you can some kind of transformation of a tan curve, not this half line: http://www10.wolframalpha.com/input/?i=arg[z-i]%3Dpi%2F2

*shrugs*

I can't work out quite what it's doing there. Wolfram Alpha is pretty new... and to be fair, it does give the solution as Re(z) = 0, Im(z) > 1 below. :p:

Reply 16

Harry1W

OP

4

Thanks a lot! Really helpful stuff. :smile:

Reply 17

Arun_gm

Whenever I run against an indeterminate funtion on the RHS like Tan(pi/2), I prefer to skirt around the problem by resorting to the cosine function(or some other convenient one) instead.

In the present case we have
arctan((y - 1)/x)=pi/2
taking cosines of the tems in the lhs and rhs we get

cos(arctan((y - 1)/x))=0

If we construct a right angled triangle with the perpendicular =(y-1) and base=x, and we take the acute angle arctan((y - 1)/x) as theta, we can see that the hypoteneuse is sqrt(x^2+y^2-2y+1), giving us the values of cos(theta) and sin(theta) as {x/sqrt(x^2+y^2-2y+1)} and {(y-1)/sqrt(x^2+y^2-2y+1)} respectively.

we just need to put cos(theta) = 0 or sin(theta) =1, whichever gives a convenient hold on the problem, and proceed from there.
Here I took the sin(theta) route, which yields:-

y^2+1-2y=x^2+y^2-2y+1

x^2=0, which at least algebraically brings me close to the result. I then figure out the limiting conditions by graphing, and trying out values for y>0, y<0, y=0 etc

Reply 18

ian.slater

12

jammer

Please don't tell me this is FP1. Or I WILL purge all over my keyboard and possibly cry simultaneously.

On the MEI syllabus, it's in FP1 - example on p68.

Reply 19

username219321

0

14

It's the point of lines where the argument of z - i is equal to pi/2 - so a straight line vertically upwards from (but not including) the point corresponding to the complex number i.

Locus of arg(z-i)=pi/2

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