2022 H2 Mathematics Paper 1 Question 1

Complex Numbers

Answers

z=2+i,{z=2 + \mathrm{i},} w=−i.{w=- \mathrm{i}.}

Full solutions

iz+2w=−1(2−i)z+iw=6\begin{align} \mathrm{i}z + 2w &= - 1 \\ (2 - \mathrm{i})z + \mathrm{i}w &= 6 \end{align}
From (1),{(1),}
w=−1−iz2\begin{equation} w = \frac{- 1-\mathrm{i}z}{2} \end{equation}
Substituting (3){(3)} into (2),{(2),}
(2−i)z+i(−1−iz2)=6(4−2i)z+i(−1−iz)=12(4−2i)z−i+z=12(5−2i)z=12+i\begin{gather*} (2 - \mathrm{i})z + \mathrm{i} \left( \frac{- 1-\mathrm{i}z}{2} \right) = 6 \\ (4 - 2 \mathrm{i})z + \mathrm{i} ( -1 - \mathrm{i} z ) = 12 \\ (4 - 2 \mathrm{i})z - \mathrm{i} + z = 12 \\ (5 - 2 \mathrm{i})z = 12 + \mathrm{i} \end{gather*}
z=12+i5−2i×5+2i5+2i=60+24i+5i+2i252+22=2+i  ■\begin{align*} z &= \frac{12 + \mathrm{i}}{5 - 2 \mathrm{i}} \times \frac{5 + 2 \mathrm{i}}{5 + 2 \mathrm{i}} \\ &= \frac{60 + 24 \mathrm{i} + 5\mathrm{i} + 2 \mathrm{i}^2}{5^2 + 2^2} \\ &= 2 + \mathrm{i} \; \blacksquare \end{align*}
w=−1−i(2+i)2=−1−2i−i22=−i  ■\begin{align*} w &= \frac{-1-\mathrm{i}(2 + \mathrm{i})}{2} \\ &= \frac{-1 - 2\mathrm{i} - \mathrm{i}^2}{2} \\ &= - \mathrm{i} \; \blacksquare \end{align*}

Question Commentary

This question involves solving simultaneous equations involving complex numbers. The approach we used above is by substitution: make w{w} the subject and substitute into the other equation to solve for z{z} by making z{z} the subject. An elimination approach will be feasible too.

Some students reach for the technique where they let z=x+yi,{z = x+y\mathrm{i},} where x{x} and y{y} are real. They then attempt to solve for x{x} and y{y} by comparing real and imaginary parts. This technique, while not incorrect, leads to much more tedious calculations.

We should strive to use simpler methods and use the comparison approach as a last resort.