Complex Locus - Modulus, Argument and Cartesian Equation Explained

[mathological]

Complex Locus - PLEASE HELP!

Homework Statement

Q1: The complex number z satisfies arg(z+3) = pi/3

(a) Sketch the locus of the point P in the Argand diagram which represents z (DONE)

(b) Find the modulus and argument of z when z takes its least value. (STUCK)[/color]

(c) Hence represent z in a + ib form. (STUCK)[/color]

Homework Equations

arg(z+3) = pi/3

The Attempt at a Solution

arctan[y/(x+3)] = pi/3

y/(x+3) = sqrt(3)

y = sqrt(3)x + 3*sqrt(3)

I can't do part (b) and hence (c) of Q1.

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Homework Statement

Q2: If z is any complex number such that |z| = 1, show using Argand diagram or otherwise that:

Homework Equations

(a) 1 <= |z+2| <= 3 (STUCK)[/color]

(b) -pi/6 <= arg(z+2) <= pi/6 (STUCK)[/color]

The Attempt at a Solution

I have sketched both parts (a) which is the region between the circles centred at (-2,0) and radii 1 and 3 and (b).

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Homework Statement

Q3: The complex number z = x+iy, x and y are real, such that |z - i| = Im(z).

(a) Show that the point representing z has a Cartesian Equation y = 1/2(x^2 + 1). Sketch the locus. (DONE)

(b) Find the gradient of the tangent to this curve which passes through the origin. Hence find the set of possible values of the principal argument of z. (STUCK)[/color]

Homework Equations

|z - i| = Im(z)

y = 1/2(x^2 + 1)

The Attempt at a Solution

for part (a)

sqrt[x² + (y-1)²] = y
x² + y² - 2y + 1 = y²
therefore, y = 1/2(x²+1)

Part (b) - NO IDEA!

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Thanks for your help!

 

[Fredrik]

(b) Find the modulus and argument of z when z takes its least value. (STUCK)[/color]

There's no ordering relation on the complex numbers. Do you mean when |z| takes its least value? In that case, you have an equation of the form y=kx+m, and you know that z=x+iy. You get |z| (as a function of x) from that. Then you find the minimum by setting its derivative =0, and so on.

Homework Statement

Q2: If z is any complex number such that |z| = 1, show using Argand diagram or otherwise that:

Homework Equations

(a) 1 <= |z+2| <= 3 (STUCK)[/color]

When it says you can use a diagram, you really just have to draw the picture and point at the details that make it obvious. If z is on the unit circle, then z+2 is on a circle with the same radius but moved 2 units to the right. So its left edge is at 1 and its right edge at 3 (because the original circle had the edges at -1 and +1). For a more rigorous solution, use the triangle inequality and the...uh...I don't know what it's called, but it's similar to the triangle inequality, but with a minus sign.

(b) -pi/6 <= arg(z+2) <= pi/6 (STUCK)[/color]

The Attempt at a Solution

I have sketched both parts (a) which is the region between the circles centred at (-2,0) and radii 1 and 3 and (b).

z+2 is on the circle with radius 1 centered at 2, not -2. So draw that circle, and then draw two lines from 0 that are tangent to the circle (above and below). Can you figure out the angles they make with the real axis?

 

[mathological]

z+2 is on the circle with radius 1 centered at 2, not -2. So draw that circle, and then draw two lines from 0 that are tangent to the circle (above and below). Can you figure out the angles they make with the real axis?

Thanks Fredrik, I was actually taking 1 <= |z+2| <= 3 as the region between circles centered at (-2,0) and with radii 1 unit and 3 units.

I see where I went wrong. But just one thing, why isn't |z+2| a circle centred at (-2,0)?

 

[mathological]

There's no ordering relation on the complex numbers. Do you mean when |z| takes its least value?

Fredrik, you are absolutely correct.

Q1 (b) is supposed to read:

(b) Find the modulus and argument of z when |z| takes its least value.

In that case, you have an equation of the form y=kx+m, and you know that z=x+iy. You get |z| (as a function of x) from that. Then you find the minimum by setting its derivative =0, and so on.

What will be the value of k and m (I mean how do you obtain these)?

 

[Fredrik]

But just one thing, why isn't |z+2| a circle centred at (-2,0)?

|z+2| is just a number. If you're thinking of the equation |z+2|=1, it defines a circle with radius 1 centered at -2, but you were given the equation |z|=1. The problem asks you to show that the two inequalities you were given are satisfied by each z on that circle.

What will be the value of k and m (I mean how do you obtain these)?

You already did that in your first post. I drew a picture and saw that the line forms a right triangle with the two axes. The bottom side has length 3, and the right side has length m, so

k=\frac{m}{3}=\tan\left(\frac{\pi}{3}\right)

and therefore

y=kx+m=k(x+3)=\tan\left(\frac{\pi}{3}\right)(x+3)

 

[mathological]

|z+2| is just a number. If you're thinking of the equation |z+2|=1, it defines a circle with radius 1 centered at -2, but you were given the equation |z|=1. The problem asks you to show that the two inequalities you were given are satisfied by each z on that circle.


You already did that in your first post. I drew a picture and saw that the line forms a right triangle with the two axes. The bottom side has length 3, and the right side has length m, so

k=\frac{m}{3}=\tan\left(\frac{\pi}{3}\right)

and therefore

y=kx+m=k(x+3)=\tan\left(\frac{\pi}{3}\right)(x+3)

Thanks Fredrik. Yeah it was a bit stupid of me to ask what k and m were.