Question 13
Use $z=a+bi$ and $w=c+di$, where $a,b,c,d\in\mathbb{R}$, to prove
[6 marks]The distance between two complex numbers
One pair of numbers is drawn on the Argand diagram, and both sides of the identity are worked out for that pair.
One example is a check, and it doesn’t prove anything. The proof has to work with $a$, $b$, $c$ and $d$. If you know the cosine rule, the triangle shows why it works: $\operatorname{Re}\left(z\overline{w}\right)=|z||w|\cos\theta$.
Follow each line of the proof with numbers
Choose values for $a$, $b$, $c$ and $d$. Each readout is one line of the proof with your numbers in place, and the last two always match.
Black: $z$. Blue: $w$. Orange: the arrow from $w$ to $z$, which is $z-w$.
With $z=0$ or $w=0$ there is no triangle, and the identity still holds. Both sides equal the modulus squared of the other number.
Here $\operatorname{Re}\left(z\overline{w}\right)=0$, so $z$ and $w$ are perpendicular and the identity is Pythagoras’ theorem: $|z-w|^2=|z|^2+|w|^2$.
Here $\operatorname{Re}\left(z\overline{w}\right)>0$, so the angle between $z$ and $w$ is less than $90^\circ$ and $|z-w|^2$ is less than $|z|^2+|w|^2$.
Here $\operatorname{Re}\left(z\overline{w}\right)<0$, so the angle between $z$ and $w$ is more than $90^\circ$ and $|z-w|^2$ is more than $|z|^2+|w|^2$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA sample response, which works each side separately until they match.
Start with the left-hand side. Substitute $z=a+bi$ and $w=c+di$, then collect the real parts and the imaginary parts.
$$\begin{aligned}\text{LHS}&=|z-w|^2\\&=|(a+bi)-(c+di)|^2\\&=|(a-c)+(b-d)i|^2\end{aligned}$$
Marker: correctly expresses $z-w$ in terms of $a$, $b$, $c$ and $d$ in Cartesian form
Since $a-c$ and $b-d$ are real, $|x+yi|^2=x^2+y^2$ applies. The square removes the square root in the modulus. Then expand both brackets.
$$\begin{aligned}&=(a-c)^2+(b-d)^2\\&=a^2-2ac+c^2\\&\qquad+b^2-2bd+d^2\end{aligned}$$
This mark can be implied by later working, and follow-through applies if Step 1 had a slip.
Marker: expresses $|z-w|^2$ in terms of $a$, $b$, $c$ and $d$ in expanded form
Now work on the right-hand side, $|z|^2+|w|^2-2\operatorname{Re}\left(z\overline{w}\right)$, one piece at a time. The first two terms are moduli squared, so each becomes a sum of two squares.
$$\begin{aligned}&|z|^2+|w|^2\\&\quad=|a+bi|^2+|c+di|^2\\&\quad=a^2+b^2+c^2+d^2\end{aligned}$$
Marker: correctly expresses $|z|^2+|w|^2$ in terms of $a$, $b$, $c$ and $d$ in expanded form
The conjugate of $w$ is $\overline{w}=c-di$. Expand the product, use $i^2=-1$, and group the real and imaginary parts.
$$\begin{aligned}z\overline{w}&=(a+bi)(c-di)\\&=ac-adi+bci-bdi^2\\&=(ac+bd)+(bc-ad)i\end{aligned}$$
So $\operatorname{Re}\left(z\overline{w}\right)=ac+bd$.
Marker: correctly expresses $\left(z\overline{w}\right)$ in terms of $a$, $b$, $c$ and $d$
Put the pieces from Steps 3 and 4 into the right-hand side, expand, and reorder the terms to match the left-hand side from Step 2.
$$\begin{aligned}\text{RHS}&=|z|^2+|w|^2\\&\qquad-2\operatorname{Re}\left(z\overline{w}\right)\\&=a^2+b^2+c^2+d^2\\&\qquad-2(ac+bd)\\&=a^2+b^2+c^2+d^2\\&\qquad-2ac-2bd\\&=a^2-2ac+c^2\\&\qquad+b^2-2bd+d^2\end{aligned}$$
This mark needs the supporting working of a proof. If an earlier slip still lets you finish the proof, follow-through applies.
Marker: completes proof
The last mark is for how the proof is set out. The marking guide’s key steps include suitable complex number notation and correct use of equals signs. Write $\text{LHS}=$ and $\text{RHS}=$ at the start of each side, so every line is equal to the one above it. Keep the modulus bars and $\operatorname{Re}$ until the step that evaluates them, and finish with a statement.
Here is the whole proof set out in full.
Let $z=a+bi$ and $w=c+di$, where $a,b,c,d\in\mathbb{R}$. Then $\overline{w}=c-di$.
$$\begin{aligned}z\overline{w}&=(a+bi)(c-di)\\&=(ac+bd)+(bc-ad)i\end{aligned}$$
So $\operatorname{Re}\left(z\overline{w}\right)=ac+bd$.
$$\begin{aligned}\text{LHS}&=|z-w|^2\\&=|(a-c)+(b-d)i|^2\\&=(a-c)^2+(b-d)^2\\&=a^2-2ac+c^2\\&\qquad+b^2-2bd+d^2\end{aligned}$$
$$\begin{aligned}\text{RHS}&=|z|^2+|w|^2\\&\qquad-2\operatorname{Re}\left(z\overline{w}\right)\\&=a^2+b^2+c^2+d^2\\&\qquad-2(ac+bd)\\&=a^2-2ac+c^2\\&\qquad+b^2-2bd+d^2\\&=\text{LHS}\end{aligned}$$
Since $\text{LHS}=\text{RHS}$, the identity is proved.
The QCAA accept other routes, such as working from one side or both sides to a suitable intermediate expression.
Marker: shows logical organisation, communicating key steps
Each phrase of the question gave you something
The question tells you which form to use, and every line after that is a modulus or an expansion. Proving an identity about $|z\pm w|^2$ in Cartesian form is a standard question type, and four of the six marks are for writing each piece in terms of $a$, $b$, $c$ and $d$.
If I had to guess where you might lose marks, it would be the conjugate. $z\overline{w}$ is $(a+bi)(c-di)$. If you multiply by $c+di$ instead, the real part comes out as $ac-bd$, and the right-hand side ends in $+2bd$ when the left has $-2bd$. The other easy loss is the sixth mark, from a single chain of equals signs that jumps between the two sides.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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