# Argument of Complex Numbers with unknown values

Find k: $$\arg (k - 4 + (5k + 1)i) = 2π/3$$

The argument of the complex number k is given.

Argument Visualisation

The real component of the complex number is NEGATIVE

The imaginary component of the complex number is POSITIVE

as the Argument is 2π/3

I constructed a triangle to show this. Triangle

and Solved for k incorrect method ,But this is incorrect i know this because when i substitute it back into the expression, it produces a complex number in a different quadrant(both components real and imaginary are negative)

correct method Is given as the correct method.

I think it might be because you can't have negative distances? so a negative k might impact the tan function.

I don't understand why my method is incorrect. And the other method seems counter productive as your using 2π/3 for tan = Opposite/Adjacent. Is this just a rule i should remember?

EDIT: Original Method

• Fix your "incorrect method": $\tan\frac\pi3=\frac{5k+1}{4-k}$, not $\frac{5k+1}{k-4}$. Commented Aug 14 at 20:13
• The formula is arctan(b/a), I'm unsure how that is incorrect? Appreciate the response Commented Aug 14 at 20:18
• See the original question. $$\tan(\frac{2π}{3})=\frac{5k+1}{k-4}$$ but $$\tan(\frac{2π}{3})=-\tan(\fracπ3)$$. This follows what Anne said
– Gwen
Commented Aug 14 at 20:22
• $k-4<0$. In your picture, the length of the horizontal segment is $4-k$. Commented Aug 14 at 20:25
• The correct answer $k = \frac{-17 + 21\sqrt{3}}{22}$ (as well as the incorrect answer) should be included in the question. I know it's in the images, but we don't like to hide essential information behind links.
– Dan
Commented Aug 14 at 22:11

The triangle construction will work as long as you are consistent about the lengths, and how they are related to trigonometric functions of associated angles in or around the triangle.

As you correctly noted, a complex number with argument $$2\pi/3$$ will be located in quadrant II, with a negative real part and a positive imaginary part. Since the complex number of interest is given by $$(k-4) + (5k+1)i$$, this tells us that $$k - 4 < 0 \quad \text{and} \quad 5k + 1 > 0.$$ If we express this as a single inequality, we obtain $$-\frac{1}{5} < k < 4. \tag{1}$$ So any solution we find must satisfy $$(1)$$.

Next, if we draw the vector for this complex number in an Argand diagram (as you have done in the first image), then we could label the triangle formed by the projection of the vector onto the real axis as follows: the leg corresponding to the real part would have magnitude $$|k-4| = 4-k$$, and the other leg corresponding to the imaginary part would have magnitude $$|5k+1| = 5k+1$$. (Not sure why you changed your variable name to $$n$$ instead of $$k$$ but it's not important.)

When we write the lengths of the triangle's sides this way, we are talking about the actual positive distance between its vertices. Then, the internal angle is $$\pi/3$$ as you correctly wrote, for which the tangent corresponds to the ratio of opposite side to the adjacent side; i.e., $$\tan \frac{\pi}{3} = \frac{5k+1}{4-k}. \tag{2}$$ Solving $$(2)$$ for $$k$$ and discarding any values of $$k$$ that do not satisfy $$(1)$$ will give the desired solution.

Alternatively, we can consider the lengths of the triangle as signed lengths--but in doing so, the angle we must consider is the supplement of the internal angle of the triangle that was drawn. In other words, the argument is $$2\pi/3$$, and the tangent of $$2\pi/3$$ will be equal to the ratio of the signed distances of the opposite leg to the adjacent leg: $$\tan \frac{2\pi}{3} = \frac{5k+1}{k-4}. \tag{3}$$ Equations $$(2)$$ and $$(3)$$ are equivalent and yield the same solution set. But you must not mix them up; e.g., writing $$\tan \frac{\pi}{3} = \frac{5k+1}{k-4} \tag{4}$$ is incorrect because when using signed distances (which may be negative), the trigonometric angle corresponds to the total angle measured counterclockwise from the positive $$x$$-axis. If this is confusing, then I recommend converting all signed distances into magnitudes, so that they are always nonnegative, and then computing trigonometric functions of angles according to their internal angle measures. But if you intend to study vector calculus, physics, and other higher-level mathematics beyond trigonometry, it is critically important to understand how to work with signed distance, which should be addressed as a topic in trigonometry.

Find $$~k ~: \arg[ ~(k - 4) + i(5k + 1) ~] = 2\pi/3.$$

This answer is much more longwinded than it would be if I employed the tangent function. However, from "An Introduction To Complex Function Theory" (Bruce Palka) - Chapter 1, I have learned a method that uses the sine and cosine functions, and avoids any confusion.

• Given $$~z = x + iy \neq 0 + i0.$$

• Compute $$~\displaystyle r = \sqrt{x^2 + y^2} \implies r > 0.$$

• Compute $$~\theta~$$ as the unique angle, within a modulus of $$~2\pi,~$$
such that $$~\displaystyle \cos(\theta) = \frac{x}{r}, ~\sin(\theta) = \frac{y}{r}.$$

Note that by the definition of $$~r,~$$ this implies that
$$\cos^2(\theta) + \sin^2(\theta) = 1.$$

A common convention, which I will follow in this answer, is to set $$~\theta~$$ as the unique satisfying angle, in the range $$~-\pi < \theta \leq \pi.~$$

So, all of the above groundwork will serve as the basis for attacking the problem.

Here, you are given that the corresponding angle, that represents $$~z = x + iy~$$ is $$~2\pi/3.$$

This implies that $$~\displaystyle \cos(\theta) = \frac{-1}{2}, ~\sin(\theta) = \frac{\sqrt{3}}{2}.$$

By the premise, you have that

• $$x = k - 4.$$

• $$y = 5k + 1.$$

This implies that

$$r = \sqrt{(k-4)^2 + (5k+1)^2}$$

$$= \sqrt{(k^2 - 8k + 16) + (25k^2 + 10k + 1)}$$

$$= \sqrt{26k^2 + 2k + 17}.$$

Therefore,

$$\frac{-1}{2} = \frac{k-4}{\sqrt{26k^2 + 2k + 17}} \tag1$$

and

$$\frac{\sqrt{3}}{2} = \frac{5k+1}{\sqrt{26k^2 + 2k + 17}}. \tag2$$

At this point, the goal is to find all values of $$~k~$$ (if any) that satisfy both (1) and (2) above. Any such satisfying value of $$~k~$$ must be construed to be a satisfying answer.

To facilitate the computations, I will compute $$~\cos^2(\theta)~$$ and $$~\sin^2(\theta),~$$ based on (1) and (2) above. This will identify all possible candidate values of $$~k,~$$ (if any). Then, each candidate value will have to be individually scrutinized to see if it satisfies both (1) and (2) above.

So,

$$\frac{1}{4} = \frac{k^2 - 8k + 16}{26k^2 + 2k + 17} \tag3$$

and

$$\frac{3}{4} = \frac{25k^2 + 10k + 1}{26k^2 + 2k + 17}. \tag4$$

The easiest way to proceed is to recognize that while (3) and (4) share the same RHS denominator, you must have the RHS numerator of (3) is $$~1/3~$$ of the RHS numerator of (4).

Therefore,

$$3 \times (k^2 - 8k + 16) = (25k^2 + 10k + 1) \implies$$

$$3k^2 - 24k + 48 = 25k^2 + 10k + 1 \implies$$

$$22k^2 + 34k - 47 = 0 \implies$$

$$k = \frac{1}{44} \left[ ~-34 \pm \sqrt{1156 + 4136} ~\right]$$

$$= \frac{1}{44} \left[ ~-34 \pm \sqrt{5292} ~\right]$$

$$= \frac{1}{44} \left[ ~-34 \pm \sqrt{(42)^2 \times 3} ~\right]$$

$$= \frac{1}{44} \left[ ~-34 \pm 42\sqrt{3} ~\right]. \tag5$$

So, the two values in (5) above represent the candidate values for $$~k.$$

From (1) above, you must have that $$~k - 4 < 0,~$$ and from (2) above, you must have that $$~(5k+1) > 0.$$

So, you must have that $$~-1/5 < k < 4.~$$

Note that $$~\displaystyle 42\sqrt{3} \approx 72.75,~$$ so you can immediately reject the candidate value of

$$\frac{1}{44} \left[ ~-34 - 42\sqrt{3} ~\right].$$

Further, the candidate value of

$$\frac{1}{44} \left[ ~-34 + 42\sqrt{3} ~\right] \approx \frac{38.75}{44}, \tag6$$

so this candidate value is in range.

It only remains to verify that the candidate value of $$~k~$$ in (6) above does in fact satisfy the original constraints represented by (1) and (2). By collectively examining the constraints represented by (1), (2), (3), and (4), walking through the computations that followed (3) and (4), and noting that

$$(k-4)^2 + (5k+1)^2 = 26k^2 + 2k + 17,$$

you can deduce that the candidate value of $$~k~$$ in (6) above does satisfy the original constraints in (1) and (2).

$$\underline{\text{Addendum}}$$

This addendum is beyond the scope of the original problem, which may be regarded as
"analytical geometry meets complex analysis", and so this addendum may well initially confuse the original poster.

In analytical geometry, the domain of the sine and cosine functions are angles, while in real analysis (AKA calculus), the domain of the sine and cosine functions are arc lengths of the unit circle (i.e. circle of radius 1).

Normally, complex analysis is considered an extension of real analysis, so normally, in complex analysis, the domain of the sine and cosine functions are arc lengths of the unit circle.

However, this specific problem is somewhat unusual in that it actually represents
"analytical geometry meets complex analysis".

That is, none of the real analysis extensions into complex analysis are relevant here. For example, this problem does not involve any differentiation, integration, or taylor series.

So, in this somewhat unusual problem, the domain of the sine and cosine functions may (harmlessly, and for the sake of simplicity) be regarded as angles.

Here, the unit circle arc length of $$~2\pi/3~$$ is represented by the angle $$~120^\circ.$$