How do I sketch the locus of |z - 5-3i | = 3 on an Argand Diagram?

First, we use the idea that a complex number z can be written in terms of its real and imaginary parts, i.e. z = x+iy, to write our expression as:

 

| x+ iy -5 - 3i | = 3

Next, we can group the real and imaginary parts of the above expression, giving us:

| (x-5) + i(y -3) | = 3

 

Now that the expression is in the form a+ib, we can use that the modulus of a complex number is the square root of (a2 + b2), to write our expression as:

[ (x-5)2 + (y-3)]1/2 = 3

 

Finally, by squaring both sides of the equation, we get:

 

(x-5)2 + (y-3) = 32

 

This sort of expression should look familiar to you; it's the standard equation for a circle!  So our final plot on our Argand diagram is of a circle center (5,3) with a radius of 3. By extending the ideas we've considered in this example, it follows that the expression |z- z1| = r represents a circle centered at z1 = x1 + iy1, with a radius r

GM
Answered by Gyen ming A. Further Mathematics tutor

27487 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

Let I(n) = integral from 1 to e of (ln(x)^n)/(x^2) dx where n is a natural number. Firstly find I(0). Show that I(n) = -(1/e) + n*I(n-1). Using this formula find I(1).


Prove by induction that the sum from r=1 to n of (2r-1) is equal to n^2.


A spring with a spring constant k is connected to the ceiling. First a weight of mass m is connected to the spring. Deduce the new equilibrium position of the spring, find its equation of motion and hence deduce its frequency f.


Prove that 1+4+9+...+n^2 = n(n+1)(2n+1)/6.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning