# Need help with complex numbers on an Argand diagram problem

Going through some complex number work for A-Level Further Maths and I have come across a question that I have had a crack at but the mark scheme is very limited so doesn't look at the method I tried to use, and I don't really understand how they tried to approach it.

Question

In an Argand diagram, the complex numbers $0, z, ze^{i\pi/6}$ are represented by the points O, A and B respectively.

i) Sketch a possible Argand diagram showing OAB. Show that the triangle is isoceles and state the size of the angle AOB.

(I was okay with this first bit)

ii) Complex numbers $1+i , 5+2i$ are represented by C and D. Complex number $w$ is represented by E such that $CD=CE$ and angle $DCE =\pi /6$ .

Calculate possible values of $w$, giving answers exactly in form $a+bi$ .

What I attempted to do was to firstly draw the triangle out again, as it was similar to the first part. I then tried to treat C as the origin so worked out that $D= 4+i$ and $E= (a-1) + (b-1)i$.

I worked out the distance between $CD= \sqrt{17}$ so tried to do $\sqrt{(a-1)^{2} +(b-1)^{2} } =\sqrt{17}$ I then worked out that $tan^{-1}(1/4)$ to find the length of CD and added $\pi /6$ to find the argument of E treating C as the origin. Then subbed in $b/a = tanANS$ and tried to solve simultaneously with my last equation.

This gave me the wrong answer. Is this approach invalid? How would I otherwise go about this problem? Thanks in advance for any advice, I reallly appreciate it! :)

edit complete workings

$$tan^{-1} (1/4) = 0.2498\\ =\pi/6 =0.7686\\ argE =0.7686\\ b/a = tan0.7686\\ =0.9667, 2.5375\\ b= 0.9667a, 2.5475a\\ (0.997b-1)^{2} + (b-1)^{2} =17\\$$

gave up here as the question says exact answers and by this point it looks like something has probably gone wrong.

MARK SCHEME ANSWERS

$$w= (1+i)+((5+2i)-(1+i))e^{\pm i\pi/6} \\ w+ 1/2 +2\sqrt{3} +(3+1/2\sqrt{3})i\\ or 3/2 +2\sqrt{3} +(-1+1/2\sqrt{3})i\\ alternative \\ CE=(a,b), CD=(4,1)\\ CE*CD =17cos{\pi/6}, CE^{2} =17 \\ 4a +b = 17\sqrt{3}/2, a^2 + b^2 =17 \\ Obtain~3-term~quadratic~in~one~variable~and~solve \\ (a,b) =(\sqrt{3} \pm 1/2 , 1/2 \sqrt{3} \mp 2)$$

(also sorry about the mildly dodgy LaTex, I'm not that used to it yet! )

• If you include your full working - especially the final answer you got and the mark scheme's answer it's easier for us to tell where your mistake is. May 13, 2018 at 14:35
• Updated! Thanks :) May 13, 2018 at 15:12

## 1 Answer

The angle $AOB$ is $\pi/6$, so the other two angles of the triangles are $$\frac{1}{2}\left(\pi-\frac{\pi}{6}\right)=\frac{5\pi}{12}$$ The triangle is isosceles because two sides have length $|z|$.

If you consider $z=(5+2i)-(1+i)=4+i$, then $w$ can have the form $$w=ze^{\pi/6}+(1+i)$$ or $ze^{-\pi/6}+(1+i)$.

• Thank you! I think the question also wants me to work out z as well? How would you say I would go about this? May 13, 2018 at 15:20
• @FlatWhiteVinyl Just do the math with $e^{\pi/6}=\frac{\sqrt{3}}{2}+\frac{1}{2}i$. May 13, 2018 at 15:22