# A question about complex numbers

1. Oct 13, 2015

### Calpalned

1. The problem statement, all variables and given/known data

I don't understand example 2. For part a, I got a slightly different answer.

2. Relevant equations
see picture

3. The attempt at a solution
$|z-1|=2=\sqrt{x^2+y^2-1}$
$4=x^2+y^2-1 \neq (x-1)^2+y^2$

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2. Oct 13, 2015

### Staff: Mentor

But the -1 is real, isn't it? Why would it affect the y (imaginary) axis? I may not understand the question well enough, though.

EDIT -- Well, that math quote didn't work so well, eh?

3. Oct 14, 2015

### SteamKing

Staff Emeritus
Expressing z into its real and imaginary parts:
z = x + iy

We can also create another complex number, w, such that:

w = u + iv

We can also specify that

u = x - 1
v = y

So that

w = u + iv = (x-1) + iy = z - 1

If we let the modulus of (z - 1) = 2, then

|w| = |z - 1| = 2

then by the definition of the modulus of a complex number, we get

|w| = (u2 + v2)1/2 = 2

Squaring both sides and substituting for u and v,

u2 + v2 = 4 or

(x - 1)2 + y2 = 4

which is the equation of a circle of radius = 2 centered at (1, 0).

4. Oct 14, 2015

### Staff: Mentor

Geometrically, the equation |z - 1| = 2 represents all of the points z in the complex plane that are 2 units away from 1; i.e., 1 + 0i. The geometry here should suggest that we're dealing with a circle of radius 2, centered at (1, 0).

For the algebra, a slightly different tack from the one SteamKing took...
$|z - 1| = 2$
$\Rightarrow |x + yi - 1| = 2$
$\Rightarrow \sqrt{ [(x - 1) + yi] \cdot [(x - 1) - yi]} = 2$ (using the fact that $|z| = \sqrt{z \cdot \bar{z}}$)
$\Rightarrow \sqrt{(x - 1)^2 + y^2} = 2$
$\Rightarrow (x - 1)^2 + y^2 = 4$

5. Oct 14, 2015

### Calpalned

Given $|z-1|$ how did you decide to make the substitution of $u=x-1$? Why not $y-1$? So my method was incorrect? While it is true that $|z| = \sqrt{x^2 +y^2}$, $|z-1| \neq \sqrt{x^2 +y^2-1}$? Thank you.

Last edited by a moderator: Oct 14, 2015
6. Oct 14, 2015

### Calpalned

Ok, this method makes a lot of sense. But just to clarify, was it wrong that I approached this question with $|z-1|=\sqrt{x^2+y^2-1}$

7. Oct 14, 2015

### Staff: Mentor

Yes, because $\sqrt{x^2+y^2-1} \ne \sqrt{(x - 1)^2 + y^2}$

8. Oct 14, 2015

### Calpalned

Okay thank you

9. Oct 14, 2015

### Calpalned

It is easier for me to follow Mark44's method, but I still thank you for your response.

10. Oct 14, 2015

### Staff: Mentor

Not even close. Subtracting 1 from z doesn't translate into subtracting 1 from the stuff inside the radical. The operations involved are very non-linear.

Forget complex numbers for a moment, and consider the hypotenuse on a 3-4-5 right triangle. We have $5 = \sqrt{3^2 + 4^2}$. If we subtract 1 from 5, does that correspond to a subtraction of 1 on the quantity in the radical? IOW, does $5 - 1 = \sqrt{3^2 + 4^2 - 1}$? I.e., is $4 = \sqrt{24}$? That's essentially what you're trying to do above.