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What I did: Cross Multiply, try to expand out the mod and args, but they all seem to lead to dead end (probably I am not seeing something)

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So: you got as far as $2z - 12 - (24 + 6z)i = 0$. What can you do if you collect the $z$ terms together, that is $(2 - 6i)z - (12 + 24i) = 0$? –  Niel de Beaudrap Nov 23 '11 at 14:54

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All your equations are correct but apparently your main difficulty was how to find $z$ in the algebraic form $a+bi$ from the given equation, which you have shown is equivalent to the linear equation in $z$ $$ 2z-12-(24+6z)i=0. $$ To solve this equation add the terms in $z$ and separately the independent terms. $$ \left( 2-6i\right) z-(12+24i)=0. $$ Move the independent term to the RHS. $$ \left( 2-6i\right) z=12+24i. $$ To find $z$ divide both sides by the coefficient of $z$, multiply both numerator and denominator by the conjugate of the denominator and simplify $$ \begin{eqnarray*} z &=&\frac{12+24i}{2-6i}=\frac{6+12i}{1-3i}=\frac{\left( 6+12i\right) \left( 1+3i\right) }{\left( 1-3i\right) \left( 1+3i\right) } \\ &=&\frac{-30+30i}{10}=-3+3i. \end{eqnarray*} $$ So $a=-3,b=3$. You can easily find the modulus of $z$ $$ \left\vert z\right\vert =\sqrt{a^{2}+b^{2}}=\sqrt{(-3)^{2}+3^{2}}=3\sqrt{2}. $$ Its argument $\theta =\arg z$ is the angle in the 2nd quadrant$^1$ such that $\tan \theta =b/a$. The range of the arctangent function is $]−\pi/2, \pi[$. Then $$ \arg z=\pi+\arctan \left( \frac{b}{a}\right) =\pi+\arctan \left( \frac{3}{-3}\right) =\pi+\arctan \left( -1\right) =\pi-\frac{1}{4}\pi=\frac{3}{4}\pi . $$

As for $w$ you do not need to find its algebraic form. You can apply the properties of the modulus and argument of a complex fraction. Since $$ \left\vert \frac{2z}{w}\right\vert =\frac{\left\vert 2z\right\vert }{ \left\vert w\right\vert }=\frac{2\left\vert z\right\vert }{\left\vert w\right\vert }=\frac{6\sqrt{2}}{\left\vert w\right\vert }=3, $$ solving for $\left\vert w\right\vert $ you get $\left\vert w\right\vert = \frac{6\sqrt{2}}{3}=2\sqrt{2}$, while from the following equation $$ \arg \left( \frac{w}{z}\right) =\arg \left( w\right) -\arg \left( z\right) =\arg \left( w\right) -\frac{3}{4}\pi =-\frac{1}{8}\pi $$ you find $\arg \left( w\right) =-\pi /8+3\pi /4=5\pi /8$.

$^1$ added and corrected

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RE: Finding $\arg{z}$: I thought since the angle in in the 2nd quadrant, I should be doing $\arg{z} = \pi - \tan^{-1}{|\frac{3}{-3}|}=\pi - \frac{\pi}{4}=\frac{3\pi}{4}$? –  Jiew Meng Nov 24 '11 at 6:43
    
@jiewmeng: You are right, the angle is in the 2nq quadrant. I corrected the associated computations. –  Américo Tavares Nov 24 '11 at 14:19

If you solve $(2 - 6i)z - (12 + 24i) = 0$ you should get that :

$z=-3+3i \Rightarrow |z|=\sqrt{(-3)^2+3^2}=3\cdot \sqrt{2}$ , and

$\arg z= \arctan(\frac{3}{-3})+ \pi=\frac{7\pi}{4}$

I assume that you can do the rest of task...

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