# Math Help - Elementary transformations (FP3)

1. ## Elementary transformations (FP3)

Points $P$ and $Q$ represent the complex numbers $w$ and $z$ respectively in the Argand diagram.

If $w=\frac{1+zi}{z+i}$, $z \neq -i$ and $w=u+iv$, $z=x+iy$, express $u$ and $v$ in terms of $x$ and $y$.

Prove that when $P$ describes the portion of the imaginary axis between the points representing $-i$ and $i$, $Q$ describes the whole of the positive imaginary axis.
I don't get what the question means. Please help me understand it.

2. Originally Posted by rednest
I don't get what the question means. Please help me understand it.
Unfortunately for you, the first part means that you have

$u + iv = \frac{1 - y + ix}{x + iy + i}$

and you have to

(i) equate the real part of the left hand side with the real part of the right hand side

(ii) equate the imaginary part of the left hand side with the imaginary part of the right hand side.

3. Originally Posted by rednest
I don't get what the question means. Please help me understand it.
I hope there is a simple way out. I had to do it by solving a strange system of equations.

So before writing out the entire solution, I want to check if my answer is right.

Is
$u = \frac{xy}{x+x^2 + (1+y)^2}$

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

EDIT: So TES proved its wrong

4. Originally Posted by rednest
I don't get what the question means. Please help me understand it.

$u+iv=\frac{1+(x+iy)i}{x+iy+i}=\frac{1-y+ix}{x+(y+1)i}$

Now multiplying by the conjugate on the left side we get

$u+iv=\frac{1-y+ix}{x+(y+1)i}\cdot \left( \frac{x-(y+1)i}{x-(y+1)i} \right)=\frac{x(1-y)-(1-y)(y+1)i+x^2i-x(y+1)}{x^2-(y+1)^2}$

$u+iv=\frac{-2xy-(1+x^2-y^2)i}{x-(y+1)^2}=\frac{-2xy}{x-(y+1)^2}+\frac{-(1+x^2-y^2)}{x-(y+1)^2}i$

equating the real and immaginary parts we get

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

We are concerned with the part of the plane $-1 \le y \le 1 \\\ x=0$ (The immaginary axis between [-i,i])

$u(0,y)=\frac{-2(0)y}{0-(y+1)^2}=0 \\\ \\\ v(0,y)=\frac{-(1+0^2-y^2)}{0-(y+1)^2}=\frac{-(1-y)(1+y)}{-(y+1)^2}=\frac{1-y}{1+y}$

So $v(0,y)$ for $y \in [-1,1]$ maps to

$[0,\infty)$

Yeah!! Thanks isomorphism!!

5. Originally Posted by TheEmptySet
$u+iv=\frac{1-y+ix}{x+(y+1)i}\cdot \left( \frac{x-(y+1)i}{x-(y+1)i} \right)=\color{red}{\frac{x(1-y)+(1-y)(y+1)i+x^2i-x(y+1)}{x^2-(y+1)^2}}$
It should be $\frac{x(1-y)-(1-y)(y+1)i+x^2i+x(y+1)}{x^2+(y+1)^2}$

6. Originally Posted by Isomorphism
It should be $\frac{x(1-y)-(1-y)(y+1)i+x^2i+x(y+1)}{x^2+(y+1)^2}$
Awsome this should fix my mapping

I will fix my above post

7. Originally Posted by TheEmptySet
$u+iv=\frac{1+(x+iy)i}{x+iy+i}=\frac{1-y+ix}{x+(y+1)i}$

Now multiplying by the conjugate on the left side we get

$u+iv=\frac{1-y+ix}{x+(y+1)i}\cdot \left( \frac{x-(y+1)i}{x-(y+1)i} \right)=\color{red}{\frac{x(1-y)-(1-y)(y+1)i+x^2i-x(y+1)}{x^2-(y+1)^2}}$
Hmm shouldnt it be

$u+iv = \frac{x(1-y)-(1-y)(y+1)i+x^2i+x(y+1)}{x^2+(y+1)^2}$??

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Here is the solution anyway:

Thus $u = \frac{2x}{x^2+(y+1)^2}$ and $v =\frac{x^2 + y^2 - 1}{x^2+(y+1)^2}$.

Now P between i and -i means $1 < v < -1$ and $u = 0$.

This means $x = 0$ and $\left|\frac{x^2 + y^2 - 1}{x^2+(y+1)^2}\right| < 1$

$\left|\frac{x^2 + y^2 - 1}{x^2+(y+1)^2}\right| < 1 \Rightarrow |x^2 + y^2 - 1|< |x^2+(y+1)^2|$

Now since x = 0,

$|y^2 - 1| < |(y+1)^2|$
$|y+1| > 0$, so divide both sides by it

$|y - 1| < |y+1|$

Its easy to prove that the above inequality holds only for $y > 0$.

But this means $x=0, |v| < 1 \Rightarrow x= 0,y = \text{Im}(Q) > 0$ and thus when " describes the portion of the imaginary axis between the points representing and , describes the whole of the positive imaginary axis."

8. Originally Posted by ThePerfectHacker
We can write $g(z)=\frac{1+zi}{z+i} = \frac{1 + (z+i) - i}{z+i}$.
How is $\frac{1+zi}{z+i} = \frac{1 + (z+i) - i}{z+i}$?

9. Originally Posted by Isomorphism
How is $\frac{1+zi}{z+i} = \frac{1 + (z+i) - i}{z+i}$?
When you had 4 shots of vodka. Sorry .

10. Originally Posted by ThePerfectHacker
When you had 4 shots of vodka. Sorry .
Lol. Nice