greatest value of the arg. of a complex number


by rock.freak667
Tags: complex, greatest, number
rock.freak667
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#1
Jun24-08, 04:54 PM
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1. The problem statement, all variables and given/known data

Given the complex number,u, is given by (7+4i)/(3-2i)

Express u in the form x+iy

Sketch the locus of z such that |z-u|=2

Find the greatest value of arg(z) for points on this locus

2. Relevant equations

For z=x+iy
[tex]|z|=\sqrt{x^2+y^2}[/tex]

[tex]arg(z)=tan^{-1}(\frac{y}{x})[/tex]

3. The attempt at a solution

First part is simply 1+2i

Second part for |z-u|=2, the locus is a circle with centre (1,2) and radius 2

third part with arg(z). Not too sure on how to find this.

I would assume the largest value for the circle is pi since it is a circle.
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Defennder
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#2
Jun24-08, 10:57 PM
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Can you picture the circle on the Argand diagram? If so, then you should be able to see at which point on the circle where z is a 'vector' from the origin of the circle to a point on the circumference, would result in the greatest angle arg(z).

EDIT: I think a better word to use, as Dick has said, would be "tangent". z is tangent to the circle.
Dick
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#3
Jun24-08, 10:59 PM
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arg(z) is just the angle that a line through z and the origin makes with the x axis. The z that makes the maximum value of that angle must be the point of intersection with the circle of a line tangent to the circle starting at the origin. Seems to me a clever person could draw some right triangles and make a trig problem out of this.

rock.freak667
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#4
Jun24-08, 11:25 PM
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greatest value of the arg. of a complex number


ok so I drew the circle, see the the real axis is a tangent to the circle at (2,0) and the Im(z) intersects the circle. So the largest arg(z) would be the tangent to the circle at the part where the circle intersects the axis?

EDIT: I think I got it out....

[itex]tan \alpha = 2[/itex] and [itex] sin \theta = \frac{2}{\sqrt{5}}[/itex]

and I need to find [itex]\alpha + \theta[/itex]
Dick
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#5
Jun24-08, 11:33 PM
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The center of the circle is (1,2). It has radius 2. I don't think it's tangent to the real axis at (2,0). It's in your best interests to draw an accurate picture or you'll waste a lot of time. The z having largest arg(z) doesn't have to be on any axis. It isn't.
Dick
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#6
Jun24-08, 11:41 PM
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Quote Quote by rock.freak667 View Post
ok so I drew the circle, see the the real axis is a tangent to the circle at (2,0) and the Im(z) intersects the circle. So the largest arg(z) would be the tangent to the circle at the part where the circle intersects the axis?

EDIT: I think I got it out....

[itex]tan \alpha = 2[/itex] and [itex] sin \theta = \frac{2}{\sqrt{5}}[/itex]

and I need to find [itex]\alpha + \theta[/itex]
I think you are thinking along the right lines since you want to add two angles, but I have a problem believing everything you say.
rock.freak667
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#7
Jun24-08, 11:44 PM
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Quote Quote by Dick View Post
I think you are thinking along the right lines since you want to add two angles, but I have a problem believing everything you say.
But the real axis is a tangent at the point (1,0) right? Since the centre is (1,2) and if we go two units down (the length of the radius), it'll be (1,0) which is on the Re(z) right?
Dick
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#8
Jun24-08, 11:46 PM
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Yes, it's (1,0) not (2,0). It's all trig now.
rock.freak667
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#9
Jun24-08, 11:46 PM
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Quote Quote by Dick View Post
Yes, it's (1,0) not (2,0). It's all trig now.
ah thanks then


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