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Step-by-Step Solution
Step 1: Identify the circle passing through the given points
The points given are
• $z_1 = 3 + 4i$ (which corresponds to the point $(3,\,4)$),
• $z_2 = 4 + 3i$ (which corresponds to the point $(4,\,3)$),
• $z_3 = 0 + 5i$ (which corresponds to the point $(0,\,5)$).
Observe that all three points satisfy $x^2 + y^2 = 25$. Hence, the circle $C$ in the complex plane passing through all three points is
$$
x^2 + y^2 = 25.
$$
This is a circle of radius 5 centered at the origin.
Step 2: Determine the slope of the line through $z_2$ and $z_3$
In Cartesian form, $z_2 = (4, 3)$ and $z_3 = (0, 5)$. The slope of the line from $z_2$ to $z_3$ is
$$
m_{23} = \frac{5 - 3}{\,0 - 4\,} \;=\; \frac{2}{-4} \;=\; -\frac{1}{2}.
$$
Step 3: Find the slope of the line through $z_1$ and $z$ (perpendicular to $z_2z_3$)
If a line is perpendicular to another line with slope $-\tfrac{1}{2}$, its slope must be the negative reciprocal. Hence, the slope $m_{1z}$ must satisfy:
$$
m_{23} \times m_{1z} = -1 \quad\Longrightarrow\quad
\left(-\frac{1}{2}\right)\,m_{1z} = -1 \quad\Longrightarrow\quad
m_{1z} = 2.
$$
Step 4: Write the equation of the line through $z_1$ and $z$
Let $z = (x,\,y)$ be the required point. Since it lies on the line through $z_1 = (3,\,4)$ with slope 2, we can use the point-slope form:
$$
\frac{y - 4}{\,x - 3\,} = 2 \quad\Longrightarrow\quad
y - 4 = 2(x - 3) \quad\Longrightarrow\quad
y = 2x - 6 + 4 = 2x - 2.
$$
This line is denoted by $L :\, y = 2x - 2$.
Step 5: Find the intersection of the line with the circle
Since $z$ lies on both the circle $x^2 + y^2 = 25$ and the line $y = 2x - 2$, we substitute $y = 2x - 2$ into the circle equation:
$$
x^2 + (2x - 2)^2 = 25.
$$
Simplify:
$$
x^2 + 4x^2 - 8x + 4 = 25,
$$
$$
5x^2 - 8x + 4 - 25 = 0,
$$
$$
5x^2 - 8x - 21 = 0.
$$
Solve this quadratic:
$$
x = \frac{\,8 \pm \sqrt{(-8)^2 - 4 \cdot 5 \cdot (-21)}\,}{2 \cdot 5}
= \frac{\,8 \pm \sqrt{64 + 420}\,}{10}
= \frac{\,8 \pm \sqrt{484}\,}{10}
= \frac{\,8 \pm 22\,}{10}.
$$
Hence, the solutions for $x$ are:
$$
x_1 = \frac{\,8 + 22\,}{10} = 3,
\quad
x_2 = \frac{\,8 - 22\,}{10} = -\frac{7}{5}.
$$
Corresponding $y$ values from $y = 2x - 2$ are:
• For $x_1 = 3$: $\,y = 2(3) - 2 = 4$, giving point $(3,4)$, which is $z_1$.
• For $x_2 = -\tfrac{7}{5}$: $\,y = 2\left(-\tfrac{7}{5}\right) - 2 = -\tfrac{14}{5} - 2 = -\tfrac{14}{5} - \tfrac{10}{5} = -\tfrac{24}{5}$.
Since $z \neq z_1$, the required intersection point is
$$
\left(-\frac{7}{5},\, -\frac{24}{5}\right).
$$
Step 6: Find the argument of the complex number $z$
The complex number corresponding to the intersection point is
$$
z = -\frac{7}{5} -\, \frac{24}{5}i.
$$
Both real and imaginary parts are negative, so $z$ is in the third quadrant. Its argument is given by
$$
\arg(z) = \tan^{-1}\!\biggl(\frac{-\tfrac{24}{5}}{-\tfrac{7}{5}}\biggr),
$$
but we must account for the third quadrant. The ratio of the imaginary part to the real part in absolute value is $\frac{24}{7}$. Since it lies in the third quadrant, the principal argument is
$$
\arg(z) = -\pi + \tan^{-1}\!\biggl(\frac{24}{7}\biggr).
$$
(Equivalently, one can write $\pi + \tan^{-1}\!\bigl(\frac{24}{7}\bigr)$ depending on the convention, but the expression $-\pi + \tan^{-1}\!\bigl(\tfrac{24}{7}\bigr)$ indicates the angle is just past $-\pi$.)
Final Answer
The required argument of $z$ is
$$
\tan^{-1}\!\Bigl(\frac{24}{7}\Bigr)\;-\;\pi.
$$
