given z=x+yi and$\displaystyle w=(z-4i)/(z+3)$, where z is not -3, has no real parts, show that
x^2+y^2+3x-4y=0
what does no real parts mean?
Yes. Just keep going...
$\displaystyle \begin{aligned}
w &= \frac{x + (y - 4)i}{(x + 3) + yi} \\
&= \frac{x + (y - 4)i}{(x + 3) + yi} \times \frac{(x + 3) - yi}{(x + 3) - yi} \\
&= \frac{x(x + 3) - xyi + (y - 4)(x + 3)i + y(y - 4)}{(x + 3)^2 + y^2} \\
&= \frac{x^2 + 3x - xyi + xyi + 3yi - 4xi - 12i + y^2 - 4y}{(x + 3)^2 + y^2}
\end{aligned}$
$\displaystyle \begin{aligned}
&= \frac{x^2 + 3x - xyi + xyi + 3yi - 4xi - 12i + y^2 - 4y}{(x + 3)^2 + y^2} \\
&= \frac{x^2 + y^2 + 3x - 4y + 3yi - 4xi - 12i}{(x + 3)^2 + y^2} \\
&= \frac{(x^2 + y^2 + 3x - 4y) + (3y - 4x - 12)i}{(x + 3)^2 + y^2} \\
&= \frac{x^2 + y^2 + 3x - 4y}{(x + 3)^2 + y^2} + \left(\frac{3y - 4x - 12}{(x + 3)^2 + y^2}\right)i
\end{aligned}$
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 $\displaystyle x^2 + y^2 + 3x - 4y = 0$.
01
EDIT: Oops, sorry to let the cat out of the bag so early.
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 $\displaystyle x^2 + y^2 + 3x - 4y = 0$.
01
understood thanks a lot!
It may be easier to use this form: $\displaystyle \frac{{z - 4i}}{{z + 3}} = \frac{{z\overline z - 4i~\overline z + 3z - 12i}}{{\left| {z + 3} \right|^2}}$.
Note that $\displaystyle \left| {z + 3} \right|^2 = \left( {x + 3} \right)^2 + y^2 $ and $\displaystyle z\overline z = x^2 + y^2 $.