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complex q (1 Viewer)
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It seems like you might have made a mistake somewhere, since it appears that the correct answer is
When
So
Would the following working help with Question 8 (a) part (ii)?
No worries. I am not entirely sure about how this was done sadly, but I am hoping that someone else would be able to clarify this bit.
hogzillaAnson
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We continue the ray by substituting into the original locus equation with a free variable that defines the modulus of z.
\ \text{for some}\ r \in \mathbb{R}^{+} \\ \text{Thus,}\ |r(3+i) - (4+3i)| &= 5 \\ |(3r-4) + (r-3)i|^2 &= 25 \\ (3r-4)^2 + (r-3)^2 &= 25 \\ 10r^2 - 30r + \cancel{25} &= \cancel{25} \\ 10r(r-3) &= 0 \\ \therefore r &= 3\ [\operatorname{arg} 0\ \text{is not defined}]\\ \text{Thus,}\ z &= 9 + 3i \implies |z| = \sqrt{9^2+3^2} = 3\sqrt{10}\ \text{as required}\end{align*})
Part ii can be solved using the triangle inequality, noting specifically the conditions for equality
 + (4+3i)| \leq |z-(4+3i)| + |4+3i| = 5 + \sqrt{4^2+3^2} = 10 \\ \text{Equality occurs precisely when}\ z-(4+3i)\ &\text{and}\ 4+3i\ \text{are parallel, i.e}\ z-(4+3i) = k(4+3i)\ \text{for some}\ k \in \mathbb{R}^{+}\\ \text{Thus,}\ z &= (k+1)(4+3i)\ (k+1 > 0) \\ \therefore \operatorname{arg} z &= \operatorname{arg} 4+3i = \tan^{-1} \frac{3}{4} \end{align*})
Part ii can be solved using the triangle inequality, noting specifically the conditions for equality
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None that I am aware of. What usually happens is that some questions may be included in more than one source, for example, more than one trial paper. In your case, the question that hogzillaAnson provided further clarification for was also included in the 2020 Hurlstone Agricultural High School Mathematics Extension 2 trial exam, which you can access using thsconline.