Complex Numbers : Argand Diagram

AI Thread Summary
The discussion focuses on sketching the region R on an Argand diagram for the inequality |iz + 1 + 3i| ≤ 3. Participants explore manipulating the equation to find the correct form, ultimately determining that the inequality represents a circle centered at (-3, 1) with a radius of 3. There is confusion regarding the modulus and the transformation of the complex number, but clarity is achieved when confirming the correct center and radius for the circle. The final consensus is that the region includes all points inside and on the boundary of this circle. The discussion emphasizes the importance of correctly interpreting complex inequalities in graphical form.
Delzac
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On an Argand diagram, sketch the region R where the following inequalities are satisfied:

l iz + 1 + 3i l less than or equal to 3

How do you draw this loci?
Do i manipulate the equation?

if so i got this :

l z - ( -3 + i ) l less than or equal to 3i

But how in the world do you draw this?

And is :

( l iz + 1 + 3i l less than or equal to 3 )= ( l z* + 1 + 3i l less than or equal to 3)

If so can is it possible to draw the z* loci and relate it to z's loci.

Any help will be greatly appreciated. Thanks.
 
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z is some complex number of the form x+iy. What is the modulus of l iz + 1 + 3i l? (Hint: Simplify iz + 1 + 3i to the form A+iB and then find the modulus.)
 
If i am going to let z = x + yi

Then i will get the following results :

l (1-y) + (3 + x)i l Less than or = 3

if so, do i draw a circle with radius 3, centre ( -1, -3) ?

So how this feels wrong.
 
Delzac said:
If i am going to let z = x + yi

Then i will get the following results :

l (1-y) + (3 + x)i l Less than or = 3

if so, do i draw a circle with radius 3, centre ( -1, -3) ?

So how this feels wrong.

It should be a circle with radius 3, centre (-3,1)... what's the modulus of l (1-y) + (3 + x)i l ?
 
Delzac said:
On an Argand diagram, sketch the region R where the following inequalities are satisfied:

l iz + 1 + 3i l less than or equal to 3

How do you draw this loci?
Do i manipulate the equation?

if so i got this :

l z - ( -3 + i ) l less than or equal to 3i

You should have l z - ( -3 + i ) l less than or equal to 3. so that's just a circle (and everything inside the circle) centered at -3+i.
 
k i got it, thanks.
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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