# Math Help - Complex No.

1. ## Complex No.

mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?

2. Hello, Navesh!

It can be solved by "eyeballing" the problem if you know a particular theorem.

$\left|\frac{z - z_1}{z-z_2}\right|\: = \:k$, where $k \neq 1$ represents a circle. .How?
We have: $|z - z_1| \:=\:k|z - z_2|$

The distance $\overline{zz_1}$ is proportional to the distance $\overline{zz_2}$.

By definition, the locus of $z$ is a Circle of Apollonius.

3. Originally Posted by Navesh
mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?
I assume $Z_{1} \mbox{ and } Z_{2}\mbox{ are given}.$
Let $Z_{1}= a +ib \mbox{ and } Z_{2} = c +id$
Let $Z = x +iy$
Now, $|Z - Z_{1}| = k|Z - Z_{2}|$
This is equivalent to $(x-a)^2 + (y-b)^2 = k^2[(x-c)^2 + (y-d)^2]$
It represents a circle if k is not equal to one. Do you know why?

4. malaygoel you used LaTex code I am proud of you

Use,
Code:
\mbox{ "type you sentences here" }
To type things like,
$Z\mbox{ and }Z'\mbox{ are given}$

5. ## circle of appolonius

Originally Posted by Navesh
mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?
what do u mean by circle of appolonius?

6. Originally Posted by Navesh
mod[(z-z1)/(z-z2)] = k where k is not equal to 1 represents a circle. how?
$
\mbox{thanks for helping}
$

7. Originally Posted by malaygoel
I assume $Z_{1} \mbox{ and } Z_{2}\mbox{ are given}.$
Let $Z_{1}= a +ib \mbox{ and } Z_{2} = c +id$
Let $Z = x +iy$
Now, $|Z - Z_{1}| = k|Z - Z_{2}|$
This is equivalent to $(x-a)^2 + (y-b)^2 = k^2[(x-c)^2 + (y-d)^2]$
It represents a circle if k is not equal to one. Do you know why?
The square terms will vanish if K=1

also simplifying the eqn. we get
$
x^2(1-k^2)+y^2(1-k^2)+2x(k^2c-a)+$
$2y(k^2d-b)-k^2(c^2+d^2)=0$
which makes the constt. term +ve, hence makes the circle real. Am I OK?

8. Hello, Navesh!

also simplifying the eqn. we get
$x^2(1-k^2)+y^2(1-k^2)+2x(k^2c-a)+$ $2y(k^2d-b)-k^2(c^2+d^2) \:= \:0$ . . . not quite
which makes the constt. term +ve, hence makes the circle real. Am I OK?

We have: . $(x - a)^2+(y-b)^2\:=\:k^2[(x-c)^2+(y-d)^2]$

Expand: . $x^2 - 2ax + a^2 + y^2 - 2by + b^2\:=\:k^2y^2$ $-\, 2k^2cy + c^2k^2 + k^2y^2 - 2k^2dy + d^2k^2$

Rearrange: . $x^2 - k^2x^2 + 2ck^2x - 2ax + y^2 - k^2y^2 + 2k^2dy \,-$ $2by \:=\: c^2k^2 + d^2k^2 - a^2 - b^2$

Factor: . $(1-k^2)x^2 + 2(k^2c - a)x + (1 - k^2)y^2 + 2(k^2d - b)$ $y\:= \;k^2(c^2 + d^2) - (a^2 + b^2)$

Hence: . $x^2 + \frac{2(k^2c - a)}{1 - k^2}x + y^2 + \frac{2(k^2d - b)}{1 - k^2}y \;=\;$ $\frac{k^2(c^2+d^2) - (a^2+b^2)}{1-k^2}$

. . .Or: . $x^2 + \frac{2(a - k^2c)}{k^2 - 1}x + y^2 + \frac{2(b - k^2d)}{k^2 - 1}y \:=\:$ $\frac{(a^2 + b^2) - k^2(c^2+d^2)}{k^2 - 1}$

And we see the restrictions on the constants.

Obviously: $k \neq \pm 1$

But equally important, the right side must be positive.

Hence: . $[1]\;a^2 + b^2 > k^2(c^2 + d^2)$ and $|k| > 1$

. . . or: . $[2]\;a^2 + b^2 < k^2(c^2 + d^2)$ and $|k| < 1$

9. Originally Posted by Soroban
Hello, Navesh!

We have: . $(x - a)^2+(y-b)^2\:=\:k^2[(x-c)^2+(y-d)^2]$

Expand: . $x^2 - 2ax + a^2 + y^2 - 2by + b^2\:=\:k^2y^2$ $-\, 2k^2cy + c^2k^2 + k^2y^2 - 2k^2dy + d^2k^2$

Rearrange: . $x^2 - k^2x^2 + 2ck^2x - 2ax + y^2 - k^2y^2 + 2k^2dy \,-$ $2by \:=\: c^2k^2 + d^2k^2 - a^2 - b^2$

Factor: . $(1-k^2)x^2 + 2(k^2c - a)x + (1 - k^2)y^2 + 2(k^2d - b)$ $y\:= \;k^2(c^2 + d^2) - (a^2 + b^2)$

Hence: . $x^2 + \frac{2(k^2c - a)}{1 - k^2}x + y^2 + \frac{2(k^2d - b)}{1 - k^2}y \;=\;$ $\frac{k^2(c^2+d^2) - (a^2+b^2)}{1-k^2}$

. . .Or: . $x^2 + \frac{2(a - k^2c)}{k^2 - 1}x + y^2 + \frac{2(b - k^2d)}{k^2 - 1}y \:=\:$ $\frac{(a^2 + b^2) - k^2(c^2+d^2)}{k^2 - 1}$

And we see the restrictions on the constants.

Obviously: $k \neq \pm 1$

But equally important, the right side must be positive.

Hence: . $[1]\;a^2 + b^2 > k^2(c^2 + d^2)$ and $|k| > 1$

. . . or: . $[2]\;a^2 + b^2 < k^2(c^2 + d^2)$ and $|k| < 1$

sorry, it was a silly mistake