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 $b$?
 A: 
$k\in\Bbb R,\,w:=k+i,\,z:=-4+i5k$ and $\arg(w+z)=\frac{2\pi}{3}$

Of course $w+z=(k-4)+i(5k+1)$. Since $0\lt\frac{2\pi}{3}\le\pi$ the (principal) argument tells us it is in the second quadrant (the top left), where $\arg(x+iy)=\pi-\arctan\frac{y}{-x}$ ("$-x$" because $x$ is negative and $-\arctan$ since we begin at $\pi$, on the negative real axis, and start turning clockwise - for this quadrant).
This means $k\lt 4$ but $5k+1\gt0$, so $-\frac{1}{5}\lt k\lt 4$. Let's compute the argument:
$$\begin{align}\arg(w+z)&\overset{\color{red}{1}}{=}\pi-\arctan\frac{5k+1}{4-k}=\frac{2\pi}{3}\\&\implies\frac{\pi}{3}=\arctan\frac{5k+1}{4-k}\\&\overset{\color{red}{2}}{\implies}\sqrt{3}=\frac{5k+1}{4-k}\\&\implies\sqrt{3}(4-k)=5k+1\\&\implies(5+\sqrt{3})k=4\sqrt{3}-1\\&\implies k=\frac{4\sqrt{3}-1}{5+\sqrt{3}}\end{align}$$
$1)$: It is also worth noting that we need $k\in\Bbb R$. If $k$ had an imaginary component, then $(k-4)$ would not be the real part and $(5k+1)$ would not be the imaginary part - they'd be a mix of both.
$2)$: It is worth noting that $\arctan(x)=y$ does not mean $x=\tan(y)$ in general. However, the way that the principal argument works, and by the fact we were careful with the quadrants, we know that it will hold for our purposes.
A: 
$w = k + i$ 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 $$\cot(\arg z)=\frac{\Re (z)}{\Im (z)}\quad\text{if }\Im (z)\ne0.$$
A: It seems to me that one can simply say
$$
w+z=k+4+i(1+5k)\\
\arg(w+z)=\tan^{-1}\frac{1+5k}{k+4}=\frac{2\pi}{3}\\
\frac{1+5k}{k+4}=\tan\frac{2\pi}{3}=-\sqrt{3}\\
k=\frac{4\sqrt{3}-1}{5+\sqrt{3}}
$$
