Greatest value of the arg. of a complex number

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Homework Help Overview

The problem involves a complex number expressed as (7+4i)/(3-2i), which needs to be simplified into the form x+iy. Following this, the task is to sketch the locus of points z such that |z-u|=2, and to determine the greatest value of arg(z) for points on this locus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the simplification of the complex number and the geometric interpretation of the locus as a circle. There are questions about how to find the maximum angle arg(z) related to points on the circle, with some suggesting the use of tangents and right triangles for analysis.

Discussion Status

The discussion is active with various interpretations of the geometry involved. Participants are exploring the relationship between the circle's properties and the angles formed with the origin. Some guidance has been offered regarding the tangential relationships, but there is no consensus on the specifics of the angles or points of interest.

Contextual Notes

There are conflicting views on the tangential points of the circle with respect to the real axis, and some participants express uncertainty about the geometric setup. The center of the circle and its radius are confirmed, but the implications for arg(z) remain under discussion.

rock.freak667
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Homework Statement



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

Homework Equations



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

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

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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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.
 
Last edited:
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.
 
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]
 
Last edited:
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.
 
rock.freak667 said:
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.
 
Dick said:
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?
 
Yes, it's (1,0) not (2,0). It's all trig now.
 
Dick said:
Yes, it's (1,0) not (2,0). It's all trig now.

ah thanks then
 

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