# complex number

• Dec 10th 2008, 04:33 AM
complex number
Given that $z=x+yi$ and $w=\frac{z+8i}{z-6}$, where z is not 6 . If w is totally imaginary , show that $x^2+y^2+2x-48=0$
• Dec 10th 2008, 06:47 AM
Complex numbers
Are you sure you have copied down the question correctly?

I think the expression should be $w=\frac{z+8}{z-6}$.

You then:

• Replace $z$ by $x+iy$ in the top and bottom
• Collect together reals and imaginaries on the top and on the bottom
• Multiply top and bottom by $(x-6)+iy$, which is the conjugate of the bottom

Don't multiply everything out at this stage. Just find the real part of the top. This must be zero if w is wholly imaginary. This will then give you $x^2+y^2+2x-48=0$

• Dec 10th 2008, 07:07 AM
Soroban

Grandad caught the typo . . .

Quote:

Given: . $z\:=\:x+yi\:\text{ and }\:w=\frac{z+{\color{red}8}}{z-6}\;\text{ where }z \neq 6.$

If $w$ is totally imaginary , show that $x^2+y^2+2x-48\:=\:0$

Substituting $z \:=\:z+yi$, we have: . $w \;=\;\frac{(x+yi) + 8}{(x+yi) - 6} \;=\;\frac{(x+8) + yi}{(x-6) + yi}$

Multiply top and bottom by $(x-6)-yi$

$w \;=\;\frac{(x+8)+yi}{(x-6)+yi}\;\cdot\;{\color{blue}\frac{(x-6) - yi}{(x-6) - yi}} \;=\;\frac{(x+8)(x-6) - y(x+8)i + y(x-6)i - y^2i^2}{(x-6)^2+y^2}$

. . $= \;\frac{x^2+2x-48 + y^2 + [y(x-6) - y(x+8)]i}{(x-6)^2+y^2}$

. . $= \;\frac{(x^2+y^2+2x-48) - 14yi}{(x-6)^2+y^2}$

. . $= \;\underbrace{\frac{x^2+y^2+2x-48}{(x-6)^2+y^2}}_{\Re} - \underbrace{\frac{14yi}{(x-6)^2+y^2}i }_{\Im}$

Since $w$ is purely imaginary, $\Re \,=\,0.$

Therefore: . $\frac{x^2+y^2+2x-48}{(x-6)^2+y^2} \:=\:0 \quad\Rightarrow\quad x^2 + y^2 +2x-48 \:=\:0$

• Dec 11th 2008, 02:36 AM