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The complex numbers z1 and z2 are given by $$z_1=5+i,z_2=2-3i$$

Determine the values of the real constants $p$ and $q$ such that $$\frac{p+iq+3z_1}{p-iq+3z_2}=2i$$

My attempt, I substitute $z_1$ and $z_2$ into the equation and it becomes

$$\frac{p+iq+15+3i}{p-iq+6-9i}=2i$$

How to proceed?

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    $\begingroup$ Why don't you try multiplying the numerator and denominator of the fraction on the LHS by the complex conjugate of the denominator and see what happens. $\endgroup$ – user137731 May 25 '15 at 4:06
  • $\begingroup$ I've typo just now. Sorry $\endgroup$ – Mathxx May 25 '15 at 4:08
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Hint: Note that the denominator simplifies to $p+6-(q+9)i$. You should simplify the numerator in the same way. Now multiply both sides of the equation by $p+6-(q+9)i$, simplify, then equate the real and imaginary parts.

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  • $\begingroup$ Do you mean $-iq-9i$ ? I've typo just now. I'm sorry $\endgroup$ – Mathxx May 25 '15 at 4:08
  • $\begingroup$ Ah. Then you should write it as $(-q-9)i$ or $-(q+9)i$. $\endgroup$ – Tim Raczkowski May 25 '15 at 4:10
  • $\begingroup$ I got it which is $p=-7$ and $q=-5$. Thanks $\endgroup$ – Mathxx May 25 '15 at 4:18

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