1. ## complex number

given z=x+yi and $w=(z-4i)/(z+3)$, where z is not -3, has no real parts, show that
x^2+y^2+3x-4y=0

what does no real parts mean?

2. Originally Posted by z1llch
given z=x+yi and $w=(z-4i)/(z+3)$, where z is not -3, has no real parts, show that Mr F says: You mean that the real part is equal to zero.
x^2+y^2+3x-4y=0

what does no real parts mean?
$w = \frac{x + iy - 4i}{x + iy + 3} = \frac{x + i(y - 4)}{(x + 3) + iy}$.

Now express $w$ in the form $w = a + ib$ in the usual way. Now put $a = 0$ since Re(w) = a.

3. i multiply the denominator with its conjugate? i got a very very long equation in the end..

4. Originally Posted by z1llch
i multiply the denominator with its conjugate? Mr F says: Yes.

i got a very very long equation in the end.. Mr F says: So start digging.
..

5. Originally Posted by z1llch
i multiply the denominator with its conjugate? i got a very very long equation in the end..
Yes. Just keep going...
\begin{aligned}
w &= \frac{x + (y - 4)i}{(x + 3) + yi} \\
&= \frac{x + (y - 4)i}{(x + 3) + yi} \times \frac{(x + 3) - yi}{(x + 3) - yi} \\
&= \frac{x(x + 3) - xyi + (y - 4)(x + 3)i + y(y - 4)}{(x + 3)^2 + y^2} \\
&= \frac{x^2 + 3x - xyi + xyi + 3yi - 4xi - 12i + y^2 - 4y}{(x + 3)^2 + y^2}
\end{aligned}

\begin{aligned}
&= \frac{x^2 + 3x - xyi + xyi + 3yi - 4xi - 12i + y^2 - 4y}{(x + 3)^2 + y^2} \\
&= \frac{x^2 + y^2 + 3x - 4y + 3yi - 4xi - 12i}{(x + 3)^2 + y^2} \\
&= \frac{(x^2 + y^2 + 3x - 4y) + (3y - 4x - 12)i}{(x + 3)^2 + y^2} \\
&= \frac{x^2 + y^2 + 3x - 4y}{(x + 3)^2 + y^2} + \left(\frac{3y - 4x - 12}{(x + 3)^2 + y^2}\right)i
\end{aligned}

Remember, this is w. The first big fraction is the real part, and the second big fraction is the imaginary part. You said that w has no real part, so the numerator of the first fraction must be zero, or $x^2 + y^2 + 3x - 4y = 0$.

01

EDIT: Oops, sorry to let the cat out of the bag so early.

6. i got
$[x^2+y^2+3x-4y-4xi+3yi+12i] / [ (x+3)^2+y^2 ]$
no real parts meaning the real part is equal to zero? i'm a little bit confused.

7. Remember, this is w. The first big fraction is the real part, and the second big fraction is the imaginary part. You said that w has no real part, so the numerator of the first fraction must be zero, or $x^2 + y^2 + 3x - 4y = 0$.

01

understood thanks a lot!

8. Originally Posted by z1llch
i got
$[x^2+y^2+3x-4-4xi+3yi+12i] / [ (x+3)^2+y^2 ]$
no real parts meaning the real part is equal to zero? i'm a little bit confused.
The real part is $\frac{x^2 + y^2 + 3x - 4{\color{red}y}}{(x + 3)^2 + y^2}$ and this will be zero if the numerator is equal to zero.

9. It may be easier to use this form: $\frac{{z - 4i}}{{z + 3}} = \frac{{z\overline z - 4i~\overline z + 3z - 12i}}{{\left| {z + 3} \right|^2}}$.

Note that $\left| {z + 3} \right|^2 = \left( {x + 3} \right)^2 + y^2$ and $z\overline z = x^2 + y^2$.