Arguments

[will74]

Hey I'm a bit confused about arguments

Would the Ray arg(z-(3+i))=pi/4 for example start at 3+i making angle of pi/4 with horizontal?
If anyone has neap 2010 paper two could you please explain what the very last question is asking? Lol

Thanks

[98.40_for_sure]

It's the same as the graph of arg(z) = pi/4, then just shift the line and the open circle to (3,1). keep translations til very last to avoid mistakes.

[kyzoo]

Hey I'm a bit confused about arguments

Would the Ray arg(z-(3+i))=pi/4 for example start at 3+i making angle of pi/4 with horizontal?
If anyone has neap 2010 paper two could you please explain what the very last question is asking? Lol

Thanks

That last question is freaking hard =.= It's probably the hardest question I've ever seen in a Spesh paper (on par with Dr. He challenge questions) and it took me like 6 pages of working out to do it (I was rushing because I was almost out of time) =X.

Ok you have the complex region C: (x+5)^2 + (y-1)^2 = 2
"u" is a point on this circle

Find "u" such that the value of Arg(u+2i) is maximum, and find an expression for Arg (u+2i) as well.

The way I did it was to find a function for Arg(u+2i) in terms of y: f(y). Then I found f'(y) using the calculator, then solving y for f"(y) = 0, you get the value of "y" such that Arg(u+2i) is maximum. Sub this back into f(y) to get the maximum value of Arg(u+2i)

As for "u" itself, I found an expression for "x" in terms of y (that's how you get Arg(u+2i) in terms of y). Sub in the same value of "y" into this expression. Now you have both "x" and "y". And u = x + yi

[Mao]

Kyzoo, is the answer x=-92/17 and y=28/17 ?

[kyzoo]

x = -92/17 and y = -6/17

[jasonn93]

i thought it was y = 28/17 ?

[kyzoo]

(28/17) - 2 = (-6/17)

You have to account for the "+2i" in Arg(u+2i)

[jasonn93]

ah yes, haha i was quoting the 'u+2i' coordinate my bad!

[Mao]

Aha.

I took a different method:

Question: Find such that is maximum for where

Solution:

Define where , dropping the apostrophe from this point on. The problem is now finding maximum
The locus (circle) lies entirely in the second quadrant, thus maximum is where the ray from origin is tangent to the circle. Define this point B. Define the center of the circle C. Define the origin O. Construct the triangle CBO, by circle geometry is a right angle. Applying pythagoras with , , this gives .

Thus the point is defined by two equations:

Solving these (by expanding the first equation, substituting , then substituting since , then solving for where ) yields the answer I got. Then substitute back to to obtain

[will74]

Hey thanks guys!

I tried it again I did it like this
-define line joining 2i and point on circle where the Ray touches (giving max arg) as y=mx+2
-implicit diff to find dy/dx for circle, dy/dx=m as we look for tangent
-sub in y=mx+c to find x in terms of m
-sub x value into y=mx+c to get y in terms of m
-sub all that into dy/dx and solve m=dy/dx
- calc gives -1,-7/23
-reject -1 as we want max argument
-sub m into previous equation for x and y
I get -92/17 -6/17i but it was not fun
Then solve for arg(u+2i)

Any problems with this method? It's pretty calculator reliant :p

Any problems with doing it this way?   

[tcg93]

Hey thanks guys!

I tried it again I did it like this
-define line joining 2i and point on circle where the Ray touches (giving max arg) as y=mx+2
-implicit diff to find dy/dx for circle, dy/dx=m as we look for tangent
-sub in y=mx+c to find x in terms of m
-sub x value into y=mx+c to get y in terms of m
-sub all that into dy/dx and solve m=dy/dx
- calc gives -1,-7/23
-reject -1 as we want max argument
-sub m into previous equation for x and y
I get -92/17 -6/17i but it was not fun
Then solve for arg(u+2i)

Any problems with this method? It's pretty calculator reliant :p

Any problems with doing it this way?   

it's mx - 2...

[will74]

Haha, yeah that's what I meant sorry...typo from working
Thanks tho