1. difficult complex numbers

The complex numbers Z and W are represented, respectively, by point P(x, y) and Q(u, v) in Argand diagrams and

W = (Z+3)/(Z+1)

The point P moves around the circle with equation |Z| = 1. Find the Cartesian equation of the locus of Q. Identify the locus.

Any ideas??

Thanks

2. Originally Posted by djmccabie
The complex numbers Z and W are represented, respectively, by point P(x, y) and Q(u, v) in Argand diagrams and

W = (Z+3)/(Z+1)

The point P moves around the circle with equation |Z| = 1. Find the Cartesian equation of the locus of Q. Identify the locus.

Any ideas??

Thanks
First write the equation

$\displaystyle W=\frac{Z+3}{Z+1}$

in terms of $\displaystyle W=u+iv$ and $\displaystyle Z=x+iy$ and simplify and equate real and imaginary parts on the two sides of the equality sign.

Now use the condition $\displaystyle |Z|=1$ to deduce $\displaystyle x^2+y^2=1$ and use this to eliminate $\displaystyle x$ or $\displaystyle y$ from the equations.

Now simplify abd hopefully the equations you are left with for $\displaystyle u$ and $\displaystyle v$ in terms of $\displaystyle x$ say will make it obvious what the locus is. To get the cartesian equation eliminate $\displaystyle x$ between the two equations.

CB

3. Hi thanks for the reply, seems to be an easier method than my tutor provided except i dont understand a few things:

W=\frac{Z+3}{Z+1}

what is \frac? does that just mean write it as the fraction?

Also:
Now use the condition |Z|=1 to deduce x^2+y^2=1

Does |Z| = x^2 + Y^2 ?

I thought |Z| = root(x^2 + Y^2)

Is there any chance you could show a solution? the work has been handed in and my tutor showed me his method but i dont fully understand.

thanks

4. Originally Posted by djmccabie
Hi thanks for the reply, seems to be an easier method than my tutor provided except i dont understand a few things:

W=\frac{Z+3}{Z+1}

what is \frac? does that just mean write it as the fraction?

Also:
Now use the condition |Z|=1 to deduce x^2+y^2=1

Does |Z| = x^2 + Y^2 ?

I thought |Z| = root(x^2 + Y^2)

Is there any chance you could show a solution? the work has been handed in and my tutor showed me his method but i dont fully understand.

thanks
The frac part is just a typo... it's LaTeX code to write fractions.

If $\displaystyle |Z| = \sqrt{x^2 + y^2} = 1$

Then squaring everything gives

$\displaystyle |Z|^2 = x^2 + y^2 = 1$.

So what Captain Black wrote is correct.

5. Originally Posted by djmccabie
Hi thanks for the reply, seems to be an easier method than my tutor provided except i dont understand a few things:

W=\frac{Z+3}{Z+1}

what is \frac? does that just mean write it as the fraction?
Sorry forgot to convert code to rendered mathematical text, fixed now.

Also:
Now use the condition |Z|=1 to deduce x^2+y^2=1

Does |Z| = x^2 + Y^2 ?

I thought |Z| = root(x^2 + Y^2)
$\displaystyle |Z|=1$ if and only if $\displaystyle |Z|^2=1$, so $\displaystyle |Z|=1$ if and only if $\displaystyle x^2+y^2=1$

CB

Originally Posted by CaptainBlack
First write the equation

$\displaystyle W=\frac{Z+3}{Z+1}$

in terms of $\displaystyle W=u+iv$ and $\displaystyle Z=x+iy$ and simplify and equate real and imaginary parts on the two sides of the equality sign.
$\displaystyle W=u+iv=\frac{(x+3)+iy}{(x+1)+iy}$

................ $\displaystyle = \frac{[(x+3)+iy][(x+1)-iy]}{(x+1)^2+y^2}$

................ $\displaystyle = \frac{ [(x+3)(x+1)+y^2] +i[y(x+1)-y(x+3)]}{(x+1)^2+y^2}$

So equating real and complex parts:

$\displaystyle u=\frac{(x+3)(x+1)+y^2}{(x+1)^2+y^2}$

$\displaystyle v=\frac{-2y}{(x+1)^2+y^2}$

CB

7. Originally Posted by CaptainBlack

Now use the condition $\displaystyle |Z|=1$ to deduce $\displaystyle x^2+y^2=1$ and use this to eliminate $\displaystyle x$ or $\displaystyle y$ from the equations.
Originally Posted by CaptainBlack
So equating real and complex parts:

$\displaystyle u=\frac{(x+3)(x+1)+y^2}{(x+1)^2+y^2}$

$\displaystyle v=\frac{-2y}{(x+1)^2+y^2}$
Now as $\displaystyle x^2+y^2=1$ we can write:

$\displaystyle u=\frac{(x^2+4x+3+y^2}{(x^2+2x+1+y^2}=\frac{4(x+1) }{2(x+1)}=2$

$\displaystyle v=\frac{-2y}{2(x+1)}=\frac{-y}{x+1}$

To simplify further observe that $\displaystyle x^2+y^2=1$ defines the unit circle so we may write:

$\displaystyle x=\cos(\theta),\ y=\sin(\theta)$

and simplify using this, or just observe that $\displaystyle u$ is a constant and $\displaystyle v$ can take any real value, so the locus is the line $\displaystyle u=2$.

CB

8. Using your method how would you solve this one?

The complex numbers Z and W are represented, respectively, by points P(x,y) and Q(u, v) in argand diagrams and

W=Z^2

The point P moves along the line Y=x-1. Find the Cartesian equation of the locus of Q

thanks