FP1 further maths complex numbers question

the statement | z + a + bi | = k

corresponds to a circle with radius k and centre ( -a, -b )

z - ( -a - bi ) can be thought of as a vector from ( -a -bi ) to z... so |z - ( -a - bi )| is th length of this vector.

How would you write that in cartesian form? Because there doesn't seem to be a rule to follow each time for how to translate the modulus form into the cartesian in terms of -ve's and +ve's. I'm just doing the question |z-1-i| which solution bank says translates to (x-1)^2+(y-1)^2=25 but this does not follow the same pattern as |z-3| which is supposed to translate to (x+3)^2+y^2=4 but this doesn't follow any pattern, so how can this be? Because the -ve in the modulus sign does not consistently translate to a -ve in the cartesian equation, unless solution bank made a mistake. How do we know when to use the -ve or +ve for the cartesian equation?

|z-1-i| can be thought of as the length of a vector from ( 1, 1 ) to z

so if |z-1-i| = 5 that will be a circle of centre ( 1, 1 ) and radius 5... so

( x - 1 )² + ( y - 1 )² = 25

I still don't understand why |z+3i|=3 goes to a -3 in the cartesian equation but |z-4i|=5 also goes to a -4 in the cartesian equation?? Surely there is some t=kind of swap over as in the +ve goes to the -ve? otherwise there is no rule? So how do we know when to use the +ve or the -ve for the cartesian equation? Why does |z+1|=1 go to (x-1)^2 etc in cartesian form but |z-3|=2 go to a +ve in cartesian form? thanks

I still don't understand why |z+3i|=3 goes to a -3 in the cartesian equation but |z-4i|=5 also goes to a -4 in the cartesian equation??

Huh?

$|z-(a+b\mathbf{i})|=r$ in the complex plane means $(x-a)^2+(y-b)^2=r^2$ in the xy plane.

So $|z+3\mathbf{i}|=3$ goes to $x^2+(y-3)^2=9$

and $|z-4\mathbf{i}|= 5$ goes to $x^2+(y+4)^2=25$ - it doesn't have a -4 in there, not sure where you saw this...

Huh?

$|z-(a+b\mathbf{i})|=r$ in the complex plane means $(x-a)^2+(y-b)^2=r^2$ in the xy plane.

So $|z+3\mathbf{i}|=3$ goes to $x^2+(y-3)^2=9$

and $|z-4\mathbf{i}|= 5$ goes to $x^2+(y+4)^2=25$ - it doesn't have a -4 in there, not sure where you saw this...

I think solution bank made a mistake as that's where I was checking my answers.

What is the solution bank that you used?

Physics and maths solution bank, do you know it?

Yeah if it's for the Heinemann textbooks then those solution banks have plenty of mistakes across all modules as I've seen many other students come to the forum with them, so don't be surprised if you find another.

Huh?

$|z-(a+b\mathbf{i})|=r$ in the complex plane means $(x-a)^2+(y-b)^2=r^2$ in the xy plane.

So $|z+3\mathbf{i}|=3$ goes to $x^2+(y-3)^2=9$

and $|z-4\mathbf{i}|= 5$ goes to $x^2+(y+4)^2=25$ - it doesn't have a -4 in there, not sure where you saw this...

I think the equations of the circles are actually x^2+(y+3)^3=9 and
x^2+(y-4)^2=25, isn't it?

Oh ok thanks for letting me know.

How do you sketch the locus of |z+3|=3|z-5|?
And what is the cartesian equation of this? But mostly how do you get to the answer?
Thanks

How do you sketch the locus of |z+3|=3|z-5|?
And what is the cartesian equation of this? But mostly how do you get to the answer?
Thanks