Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$ I am working on problem 2B-11 from the book Core Pure Mathematics, Edexcel 1/AS. The question is:
The complex numbers $w$ and $z$ are given by $w = k + i$ and $z = -4 + 5ki$ where $k$ is a real constant. Given at $\arg(w + z) = \frac{2\pi}{3}$, find the exact value of $k$.
I have attempted this question, but by answer is different from the given one and I can't figure out why. My answer:
Let $w + z = a + bi$.
$a = k - 4 \\
b = 1 + 5k$
The angle of the triangle formed by the vector $w + z$ (on Argand diagram) and the x-axis is $\pi -\frac{2\pi}{3} = \frac{\pi}{3}$
$\tan\frac{\pi}{3} = \frac{b}{a} \\
\tan\frac{\pi}{3} = \frac{1 + 5k}{k - 4} \\
k\tan\frac{\pi}{3} - 4\tan\frac{\pi}{3} = 1 + 5k \\
k\tan\frac{\pi}{3} - 5k = 1 + 4\tan\frac{\pi}{3} \\
k = \frac{1 + 4\tan\frac{\pi}{3}}{\tan\frac{\pi}{3} - 5} = -2.42$
However, the given answer is:
$\frac{4\sqrt{3} - 1}{5 + \sqrt{3}} = 0.88$  where $\tan\frac{\pi}{3} = \sqrt{3}$
I have just noticed that assuming the angle is $-\tan\frac{\pi}{3}$, my answer is correct. How can you tell the quadrant of $w + z$ when the constant $k$ appears in both $a$ and $b$?


$w = k + i$ and $z = -4 + 5ki$
Let $w + z = a + bi$.
$a = k - 4 \\b =i + 5ki$

Typo: $b=1+5k.$

The angle of the triangle formed by the complex number $w + z$ is $$\pi -\frac{2\pi}{3} = \frac{\pi}{3}.$$

"The triangle formed by $w + z$" is ambiguous.
In any case, your critical error is here, where you assume that $a$ and $b$ have the same sign:

$$\tan\frac{\pi}{3} = \frac{b}{a}.$$


Edit

The angle of the triangle formed by the vector w+z (on Argand diagram)

This is no less ambiguous than the original statement!

I still do not understand however, although I made that bad assumption, how do you know the sign of $a$ and $b,$ to be able to answer the question?

Since the argument of $a+bi$ is $\dfrac23\pi,$ then $a$ is negative and $b$ positive.
So, $$\operatorname{Arg}(a+bi)=\pi-\arctan\left(\frac b{-a}\right).$$
In general, $$\tan(\arg z)=\frac{\Im (z)}{\Re (z)}\quad\text{if }\Re (z)\ne0,$$ and $$\cot(\arg z)=\frac{\Re (z)}{\Im (z)}\quad\text{if }\Im (z)\ne0.$$
