Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$

Question

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 $b$?

Answer 1

$k\in\Bbb R,\,w:=k+i,\,z:=-4+i5k$ 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.

Answer 2

$w = k + i$ 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:

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

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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 $$\cot(\arg z)=\frac{\Re (z)}{\Im (z)}\quad\text{if }\Im (z)\ne0.$$

Answer 3

It seems to me that one can simply say

$$ w+z=k+4+i(1+5k)\\ \arg(w+z)=\tan^{-1}\frac{1+5k}{k+4}=\frac{2\pi}{3}\\ \frac{1+5k}{k+4}=\tan\frac{2\pi}{3}=-\sqrt{3}\\ k=\frac{4\sqrt{3}-1}{5+\sqrt{3}} $$