Let $z$ be the complex number satisfying $|z - 5| \le 3$ and having maximum positive principal argument. Then $34 \left| \frac{5z - 12}{5iz + 16} \right|^2$ is equal to :
Region for Complex Number z
The condition $|z - 5| \le 3$ defines a closed disk in the complex plane. This disk is centered at $5$ (corresponding to the point $(5, 0)$) and has a radius of $3$. All complex numbers $z$ satisfying this condition lie within or on the boundary of this circle.
Finding z with Maximum Positive Argument
We seek the complex number $z$ within the disk $|z - 5| \le 3$ that possesses the maximum positive principal argument. The principal argument $\theta$ satisfies $-\pi < \theta \le \pi$. Since the disk is centered at $(5, 0)$ with radius $3$, all points $z = x + iy$ in the disk have $x \ge 2$. Consequently, all points are located in the first or fourth quadrant, meaning their arguments lie within the range $(-\pi/2, \pi/2)$. The maximum positive argument occurs at a point on the boundary circle, $(x-5)^2 + y^2 = 9$, where the line segment connecting the origin to $z$ is tangent to the circle.
Let $z = x + iy$. For tangency, the radius vector from the center $(5, 0)$ to $z = (x, y)$ must be perpendicular to the vector from the origin to $z = (x, y)$. The dot product of these vectors, $(x-5, y)$ and $(x, y)$, must be zero:
$x(x-5) + y^2 = 0$ $x^2 - 5x + y^2 = 0 \quad (*)$
Since $z$ is on the circle boundary, it also satisfies $(x-5)^2 + y^2 = 9$. Expanding this yields:
$x^2 - 10x + 25 + y^2 = 9 \quad (**)$
Substitute $y^2 = 5x - x^2$ from (*) into (**):
$x^2 - 10x + 25 + (5x - x^2) = 9$ $-5x + 25 = 9$ $-5x = -16$ $x = \frac{16}{5}$
Find $y$ using $y^2 = 5x - x^2$:
$y^2 = 5\left(\frac{16}{5}\right) - \left(\frac{16}{5}\right)^2 = 16 - \frac{256}{25} = \frac{400 - 256}{25} = \frac{144}{25}$ $y = \pm \frac{12}{5}$
To achieve the maximum positive argument, we select the positive value for $y$. Therefore, the complex number is:
$z = \frac{16}{5} + i \frac{12}{5}$
Calculating the Expression Value
The expression to evaluate is $34 \left| \frac{5z - 12}{5iz + 16} \right|^2$. Substitute the determined value of $z$.
Calculate the numerator $5z - 12$:
$5z - 12 = 5\left(\frac{16}{5} + i \frac{12}{5}\right) - 12 = (16 + 12i) - 12 = 4 + 12i$
Calculate the denominator $5iz + 16$:
$5iz + 16 = 5i\left(\frac{16}{5} + i \frac{12}{5}\right) + 16 = i(16 + 12i) + 16 = 16i + 12i^2 + 16$ $= 16i - 12 + 16 = 4 + 16i$
Compute the fraction:
$\frac{5z - 12}{5iz + 16} = \frac{4 + 12i}{4 + 16i} = \frac{4(1 + 3i)}{4(1 + 4i)} = \frac{1 + 3i}{1 + 4i}$
Determine the squared magnitude of this fraction:
$\left| \frac{1 + 3i}{1 + 4i} \right|^2 = \frac{|1 + 3i|^2}{|1 + 4i|^2} = \frac{1^2 + 3^2}{1^2 + 4^2} = \frac{1 + 9}{1 + 16} = \frac{10}{17}$
Finally, compute the value of the entire expression:
$34 \left| \frac{5z - 12}{5iz + 16} \right|^2 = 34 \times \frac{10}{17} = 2 \times 10 = 20$
Similar Questions
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
- The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
- Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to :
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to
Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
