arg(z+2+2i)=2pi/3 - is my book wrong?

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Mad Man
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My book tells me to sketch the locus of arg(z+2+2i)=2pi/3 but if you simplify you get arg(x+2) + (y+2)i)=2pi/3 when you x=1 y=1 you get pi/4. In fact you can do this for any value foe when x=y and you get pi/4. If not what values of x and y do you get?

In the answer it says x<-2 and y>-2 so if you use x=-3 and y=-1 you get 3/4 pi
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DFranklin
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Would you know how to sketch the locus: arg(z) = 2pi/3?
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Mad Man
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(Original post by DFranklin)
Would you know how to sketch the locus: arg(z) = 2pi/3?
yes It is a line that goes from the origin at counterclockwise angle of 2pi/3 but it starts at the origin.
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Mad Man
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why do you ask?
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Mad Man
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Can you please help
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Mad Man
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Please muttley777
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DFranklin
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(Original post by Mad Man)
yes It is a line that goes from the origin at counterclockwise angle of 2pi/3 but it starts at the origin.
Well, if arg(u) = 2pi/3, and z = u - 2 - 2i, can you see how this means that arg(z+2+2i) = 2pi/3?
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Mad Man
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please Muttley79
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Mad Man
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(Original post by DFranklin)
Well, if arg(u) = 2pi/3, and z = u - 2 - 2i, can you see how this means that arg(z+2+2i) = 2pi/3?
no can you please explain?
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Mad Man
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Oh wait, the numbers cancel out right? But is there a graph calculator that uses argand diagrams I want to visualise it please
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Mad Man
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(Original post by Mad Man)
In the answer it says x<-2 and y>-2 so if you use x=-3 and y=-1 you get 3/4 pi
Why is this the case though. What I am saying is that is there a value of x and y for z=x+iy to get arg(z+2+2i)=2pi/3?
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DFranklin
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(Original post by Mad Man)
no can you please explain?
Well z + 2 + 2i is going to equal u.

So this means that if you know the locus of u (which you described in post #3), and you shift that by -2 - 2i, you will get the locus for z.
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DFranklin
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(Original post by Mad Man)
Why is this the case though. What I am saying is that is there a value of x and y for z=x+iy to get arg(z+2+2i)=2pi/3?
Yes. Again, would you know how to find values for x and y so that if u = x+iy then arg(u) = pi/3?
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Mad Man
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(Original post by DFranklin)
Yes. Again, would you know how to find values for x and y so that if u = x+iy then arg(u) = pi/3?
Would you sub in values for x and y and use the arg button on your calculator?
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Mad Man
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(Original post by DFranklin)
Yes. Again, would you know how to find values for x and y so that if u = x+iy then arg(u) = pi/3?
Would you do tan-1(y/x)=pi/3 then y/x =tan(pi/3) but how would you solve this?
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DFranklin
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(Original post by Mad Man)
Would you sub in values for x and y and use the arg button on your calculator?
That's going to tell you what arg(x+iy) is, but it's not going to tell you how to find x and y to give a particular argument.

So, here's a basic fact you should know. The point z with |z| = R (i.e. distance from the origin = R) and argument theta is R \cos \theta + R i \sin \theta.

Edit: however, you don't need this to sketch the locus. You know the angle the line has to go in (post #3), and you know where it starts from (post #12). So you don't need equations for x and y.
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Mad Man
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(Original post by DFranklin)
Yes. Again, would you know how to find values for x and y so that if u = x+iy then arg(u) = pi/3?
(Original post by DFranklin)
That's going to tell you what arg(x+iy) is, but it's not going to tell you how to find x and y to give a particular argument.

So, here's a basic fact you should know. The point z with |z| = R (i.e. distance from the origin = R) and argument theta is R \cos \theta + R i \sin \theta.
Oh yeah I knew that, it was on the last page on my book, I'll try and work what x and y are then.
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DFranklin
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(Original post by Mad Man)
Oh yeah I knew that, it was on the last page on my book, I'll try and work what x and y are then.
You don't need to know what x and y are. (See edit to previous post).
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Mad Man
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(Original post by DFranklin)
That's going to tell you what arg(x+iy) is, but it's not going to tell you how to find x and y to give a particular argument.

So, here's a basic fact you should know. The point z with |z| = R (i.e. distance from the origin = R) and argument theta is R \cos \theta + R i \sin \theta.

Edit: however, you don't need this to sketch the locus. You know the angle the line has to go in (post #3), and you know where it starts from (post #12). So you don't need equations for x and y.
Please wait I need to show you what my book says and I want to see if you agree with it because it says you need to form an equation for x and y
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Mad Man
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(Original post by Mad Man)
Please wait I need to show you what my book says and I want to see if you agree with it because it says you need to form an equation for x and y
This is what the book says DFranklin
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