Let a circle C in complex plane pass through the points z1 = 3 + 4i, z2 = 4 + 3i and z3 = 5i. If z (

Q: Wait, how did you know the circle was centered at the origin?

Great observation! In the complex plane, the distance of a point $z = x + iy$ from the origin is given by $|z| = \sqrt{x^2 + y^2}$. If you calculate the modulus for each point: $|z_1| = \sqrt{3^2 + 4^2} = 5$, $|z_2| = \sqrt{4^2 + 3^2} = 5$, and $|z_3| = \sqrt{0^2 + 5^2} = 5$. Since all three points are at a distance of $5$ from the origin, they must lie on the circle $x^2 + y^2 = 25$.

Q: Why do we reject $x = 3$?

The problem explicitly states that $z \neq z_1$. Since $z_1 = 3 + 4i$, its x-coordinate is $3$. When we solve the intersection of the line and the circle, we get two points of intersection. One of them is $z_1$ itself, and the other is our target point $z$. Therefore, we must reject the root $x = 3$ to find the unique point $z$.

Q: Why is the argument $\tan^{-1}(y/x) - \pi$?

The principal argument of a complex number $z = x + iy$ is $\tan^{-1}(y/x)$ only in the first and fourth quadrants. In the third quadrant (where both $x$ and $y$ are negative), the angle measured from the positive real axis is $\tan^{-1}(y/x) - \pi$. Since our point $z$ has both negative coordinates, this adjustment is necessary to get the correct principal argument.

Question

Let a circle $C$ in complex plane pass through the points $z_1 = 3 + 4i$, $z_2 = 4 + 3i$ and $z_3 = 5i$. If $z (
eq z_1)$ is a point on $C$ such that the line through $z$ and $z_1$ is perpendicular to the line through $z_2$ and $z_3$, then $\arg(z)$ is equal to :

Options:

(A) $	an^{-1}\left(\frac{2}{\sqrt{5}}ight) - \pi$

(B) $	an^{-1}\left(\frac{24}{7}ight) - \pi$

(C) $	an^{-1}(3) - \pi$

(D) $	an^{-1}\left(\frac{3}{4}ight) - \pi$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the complex plane! Today, we are going to unravel a problem that might look like a standard complex number exercise, but is, at its heart, a beautiful dance of geometry.

We are given three points, $z_1 = 3 + 4i$, $z_2 = 4 + 3i$, and $z_3 = 5i$, and told they lie on a circle $C$. Our mission is to find the argument of a fourth point $z$ on this circle, defined by a specific perpendicularity condition.

### Unmasking the Circle

First, let's stop viewing these as abstract complex numbers and start seeing them as coordinates on the Cartesian plane. $z_1$ becomes $(3, 4)$, $z_2$ becomes $(4, 3)$, and $z_3$ becomes $(0, 5)$.

Now, look at their distances from the origin. We calculate the modulus for each: 
$$|z_1|^2 = 3^2 + 4^2 = 25$$
$$|z_2|^2 = 4^2 + 3^2 = 25$$
$$|z_3|^2 = 0^2 + 5^2 = 25$$

This is the 'Aha!' moment. All three points are exactly $5$ units away from the origin, meaning they all lie on the circle defined by the equation $x^2 + y^2 = 25$. The circle is centered at the origin with a radius of $5$.

### The Perpendicular Dance

Next, we need to find the line passing through $z_2$ and $z_3$. The slope $m_1$ of this line is given by the change in $y$ over the change in $x$:

$$m_1 = \frac{5 - 3}{0 - 4} = \frac{2}{-4} = -\frac{1}{2}$$

The problem tells us that the line through $z$ and $z_1$ is perpendicular to this line. In the world of coordinate geometry, two lines are perpendicular if the product of their slopes is $-1$.

So, if $m_1 = -\frac{1}{2}$, then the slope $m_2$ of our new line must be $2$.

### The Intersection

Now we have the slope $m_2 = 2$ and a point on the line, $z_1 = (3, 4)$. Using the point-slope form, $y - y_1 = m(x - x_1)$, we get $y - 4 = 2(x - 3)$, which simplifies to $y = 2x - 2$.

Since $z$ also lies on the circle $x^2 + y^2 = 25$, we substitute $y = 2x - 2$ into the circle equation:

$$x^2 + (2x - 2)^2 = 25$$

Expanding this, we get $x^2 + 4x^2 - 8x + 4 = 25$, which simplifies to the quadratic:

$$5x^2 - 8x - 21 = 0$$

Factoring this, we find $(5x + 7)(x - 3) = 0$. This gives us two possible $x$-coordinates: $x = 3$ and $x = -\frac{7}{5}$. Since the problem states $z 
eq z_1$, we reject $x = 3$ and accept $x = -\frac{7}{5}$.

### Final Calculation

With $x = -\frac{7}{5}$, we find the $y$-coordinate:

$$y = 2\left(-\frac{7}{5}ight) - 2 = -\frac{14}{5} - \frac{10}{5} = -\frac{24}{5}$$

Our point $z$ is $(-\frac{7}{5}, -\frac{24}{5})$. Both coordinates are negative, placing $z$ firmly in the third quadrant.

The principal argument $\arg(z)$ for a point in the third quadrant is calculated as $	an^{-1}(\frac{y}{x}) - \pi$. Substituting our values, we get:

$$\arg(z) = 	an^{-1}\left(\frac{-24/5}{-7/5}ight) - \pi = 	an^{-1}\left(\frac{24}{7}ight) - \pi$$

And there it is! The final answer is **$	an^{-1}(\frac{24}{7}) - \pi$**.

Select Answer:

Visualized Solution

Identify Points as Coordinates

Determine Circle Equation

Calculate Slope of $z_{2} z_{3}$

Determine Slope of $z z_{1}$

Equation of Line $z z_{1}$

Substitute Line into Circle

Solve the Quadratic Equation

Identify Point $z$

Determine the Quadrant

Calculate Final Argument

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Unmasking the Circle

The Perpendicular Dance

The Intersection

Final Calculation

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