All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
z^{2}=(1+2i)^{2}

z^{2}=1+4i-4=-3+4i

\frac{1}{z^{3}}=\frac{1}{(1+2i)(-3+4i)}

\frac{1}{z^{3}}=\frac{1}{-11-2i}\times \frac{-11+2i}{-11+2i}

\frac{1}{z^{3}}=\frac{-11+2i}{121+4}

\frac{1}{z^{3}}=\frac{-11}{125}+\frac{2i}{125}

(ii)
Using (i)
pz^{2}+\frac{q}{z^{2}}=p(-3+4i)+q(\frac{-11+2i}{125})

Since it is real, 4p+\frac{2q}{125}=0

We have q=-250p,

pz^{2}+\frac{q}{z^{2}}=-3p-(\frac{11}{125})(-250p)=19p

Personal Comments:
(i) tests students on their understanding of manipulation and also how to rationalise a complex number. I strongly encourage students to still use the calculator to check their answers.
(ii) is quite comfortable so long as students realise the meaning of being real.

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