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FP2 Complex Numbers

This question is regarding using complex numbers to locate the radius and the center of the circle.

So i have this I z-5-3i I =3 (the two end lines represent the modulus).

I am confused on how to find the Cartesian equation.

I understand we use z=x+iy, so we get I x+iy-5-3i I.= 3

We group real and imaginery, so I (x-5)+i(y-3) I =3

I do NOT understand or know what to do next. How do you remove the modulus, where does the " i " go?

The next step is (x-5)^2 + (y-3)^2 = 9

But ONCE again how does one eliminate the modulus sign and as far as i know i^2 is -1. so i don't see any signs changing.

What are the necessary step
Sorry you've not had any responses about this. :frown: Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses. :redface:

I'm going to quote in Puddles the Monkey now so she can move your thread to the right place if it's needed. :h: :yy:

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Avatar for TeeEm
TeeEm
19
Original post by john122334455
This question is regarding using complex numbers to locate the radius and the center of the circle.

So i have this I z-5-3i I =3 (the two end lines represent the modulus).

I am confused on how to find the Cartesian equation.

I understand we use z=x+iy, so we get I x+iy-5-3i I.= 3

We group real and imaginery, so I (x-5)+i(y-3) I =3

I do NOT understand or know what to do next. How do you remove the modulus, where does the " i " go?

The next step is (x-5)^2 + (y-3)^2 = 9

But ONCE again how does one eliminate the modulus sign and as far as i know i^2 is -1. so i don't see any signs changing.

What are the necessary step


|a+bi| = c
√(a2 + b2) = c (by definition)
square both sides
a2 + b2 =c2
Original post by john122334455
This question is regarding using complex numbers to locate the radius and the center of the circle.

So i have this I z-5-3i I =3 (the two end lines represent the modulus).

I am confused on how to find the Cartesian equation.

I understand we use z=x+iy, so we get I x+iy-5-3i I.= 3

We group real and imaginery, so I (x-5)+i(y-3) I =3

I do NOT understand or know what to do next. How do you remove the modulus, where does the " i " go?

The next step is (x-5)^2 + (y-3)^2 = 9

But ONCE again how does one eliminate the modulus sign and as far as i know i^2 is -1. so i don't see any signs changing.

What are the necessary step


Think of the complex number (x-5) + i(y-3) as a vector (or just a line). The modulus is the length of that line so just use Pythagoras. The i doesn't come into it.

More easily, from the first line, the pattern of the equation can be written as z - (5 + 3i) so you have a circle, centre (5,3) radius 3, with the radius coming from the 3 on the right hand side. So you can write down the Cartesian equation straight away.
Original post by TSR Jessica
Sorry you've not had any responses about this. :frown: Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses. :redface:

I'm going to quote in Puddles the Monkey now so she can move your thread to the right place if it's needed. :h: :yy:




Thank you so much i really appreciete, although after your post i tried searching for the fp2 study help forum, couldnt find it. lol. should i stop posting questions on this tread
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Avatar for Phichi
Phichi
11
Original post by john122334455
Thank you so much i really appreciete, although after your post i tried searching for the fp2 study help forum, couldnt find it. lol. should i stop posting questions on this tread


Your thread is in the Maths section (the right place), continue to post at will.

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