Engineering Mathematics – GATE PI Engineering MathematicsPractice Questions & PYQs

The Newton-Raphson method helps us find roots of an equation iteratively using the formula: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. Here, $f(x) = x^5 - 15$, so its derivative is $f'(x) = 5x^4$. Given $x_0 = 1.0$, we calculate $f(x_0) = (1.0)^5 - 15 = -14$ and $f'(x_0) = 5 \times (1.0)^4 = 5$. Plugging these into the formula for $n=0$:
$x_1 = 1.0 - \frac{-14}{5} = 1.0 + 2.8 = 3.8$.
Rounding to two decimal places, $x_1 = 3.80$.

To find the determinant of matrix $A$, we first perform the matrix multiplication.
$A = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 4 \end{bmatrix} = \begin{bmatrix} 3 \times 1 & 3 \times 4 \\ 2 \times 1 & 2 \times 4 \end{bmatrix} = \begin{bmatrix} 3 & 12 \\ 2 & 8 \end{bmatrix}$
Now, using the formula $\det(A) = ad - bc$ for a $2 \times 2$ matrix, we calculate the determinant:
$\det(A) = (3)(8) - (12)(2) = 24 - 24 = 0$
Therefore, the value of $\det(A)$ is $0$.

Let's understand how squaring a matrix affects its eigenvalues and eigenvectors.

We start with the definition of an eigenvalue $\lambda$ and its corresponding eigenvector $v$ for a matrix $A$:
$Av = \lambda v$
Now, let's multiply both sides by $A$ from the left:
$A(Av) = A(\lambda v)$
Using the associative property on the left and factoring out the scalar $\lambda$ on the right, we get:
$A^2v = \lambda (Av)$
Substitute $Av = \lambda v$ back into the equation:
$A^2v = \lambda (\lambda v)$
Simplifying the right side gives us:
$A^2v = \lambda^2 v$
This equation shows that $v$ is still an eigenvector for $A^2$, but its corresponding eigenvalue is $\lambda^2$. Therefore, when a matrix is squared, its eigenvalues get squared, but its eigenvectors are retained.

The question asks for the expansion of the vector triple product $(a \times b) \times c$. This is a direct application of a fundamental identity in vector algebra. The Vector Triple Product Identity is given by:
$(a \times b) \times c = (a \cdot c)b - (b \cdot c)a$
This identity shows that the result is a vector lying in the plane formed by vectors $a$ and $b$, and it's expressed as a linear combination of $a$ and $b$. Comparing this standard identity with the provided options, we find that Option C, which is $(a \cdot c)b - (b \cdot c)a$, exactly matches the formula.

Let's find the value of $\lim_{x\to1} \frac{x^{3-}3x+2}{x^{3-}x^{2-}x+1}$.

First, we substitute $x=1$ into the expression to check the form.
Numerator: $1^3 - 3(1) + 2 = 1 - 3 + 2 = 0$
Denominator: $1^3 - 1^2 - 1 + 1 = 1 - 1 - 1 + 1 = 0$
Since we have the indeterminate form $\frac{0}{0}$, we can use factorization.

Because $x=1$ makes both numerator and denominator zero, $(x-1)$ must be a factor of both.
Factor the Numerator: $x^3 - 3x + 2$
We can see that $x^3 - 3x + 2 = (x-1)(x^{2+}x-2)$.
Further factoring the quadratic part gives $x^{2+}x-2 = (x-1)(x+2)$.
So, the numerator becomes $(x-1)(x-1)(x+2) = (x-1)^2(x+2)$.

Factor the Denominator: $x^3 - x^2 - x + 1$
Group terms: $x^2(x-1) - 1(x-1) = (x^{2-}1)(x-1)$.
Using the difference of squares formula, $x^{2-}1 = (x-1)(x+1)$.
So, the denominator becomes $(x-1)(x+1)(x-1) = (x-1)^2(x+1)$.

Now, substitute these factored forms back into the limit expression and simplify for $x \neq 1$:
$\lim_{x\to1} \frac{(x-1)^2(x+2)}{(x-1)^2(x+1)} = \lim_{x\to1} \frac{x+2}{x+1}$
Finally, substitute $x=1$ into the simplified expression:
$\frac{1+2}{1+1} = \frac{3}{2} = 1.5$
The value of the limit is $1.5$, which, when rounded to one decimal place, is $1.5$. This falls within the range of $1.4$ to $1.6$.

For numerical integration methods, we often talk about their "order of exactness." This refers to the highest degree polynomial for which the method gives an exact result. The Trapezoidal Rule provides exact results for polynomials of degree 1 or less. Simpson's 1/3 Rule is more accurate and provides exact results for polynomials of degree 3 or less. The given function, $y = 2x + 5$, is a linear function, which is a polynomial of degree 1. Since the degree of the function (1) is less than or equal to the exactness limit for both the Trapezoidal Rule (1) and Simpson's 1/3 Rule (3), both methods will compute the integral exactly.

This is a second-order linear homogeneous ordinary differential equation (ODE) with constant coefficients: $4\frac{d^2y}{dx^2}-4 \frac{dy}{dx} + y = 0$. To find the general solution, we form the characteristic equation by replacing $\frac{d^2y}{dx^2}$ with $r^2$ and $\frac{dy}{dx}$ with $r$, which gives $4r^2 - 4r + 1 = 0$. Factoring this quadratic equation yields $(2r - 1)^2 = 0$, indicating a repeated real root $r = \frac{1}{2}$. For a repeated real root $r$, the general solution of such an ODE is $y = (c_1 + c_2x)e^{rx}$. Substituting $r = \frac{1}{2}$ into this form, we get $y = (c_1 + c_2x)e^{\frac{1}{2}x}$ or equivalently $y = (c_1 + c_2x)e^{x/2}$.

This problem asks us to find the new sample size needed to make a confidence interval narrower, while keeping the same confidence level.

The core idea is how the margin of error ($E$) of a confidence interval relates to the sample size ($n$). For a normal distribution, the margin of error is given by the formula:
$E = Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$
Here, $Z_{\alpha/2}$ is a constant for a given confidence level, and $\sigma$ is the population standard deviation, which is also constant. This means the margin of error is inversely proportional to the square root of the sample size: $E \propto \frac{1}{\sqrt{n}}$.
Consequently, the sample size ($n$) is inversely proportional to the square of the margin of error ($E^2$): $n \propto \frac{1}{E^2}$.

Let's calculate the initial margin of error ($E_1$) and the desired margin of error ($E_2$):

1. Initial Scenario:

   • Confidence Interval: $[500, 700]$
   • Mean: $600$
   • Initial Margin of Error ($E_1$): Half the width of the interval.
     $E_1 = \frac{700 - 500}{2} = \frac{200}{2} = 100$
   • Initial Sample Size ($n_1$): $1000$

2. Desired Scenario:

   • Desired Confidence Interval: $[550, 650]$
   • Desired Margin of Error ($E_2$): Half the width of the new interval.
     $E_2 = \frac{650 - 550}{2} = \frac{100}{2} = 50$

Now, using the relationship $n \propto \frac{1}{E^2}$, we can set up a ratio for the sample sizes and margins of error:
$\frac{n_2}{n_1} = \left(\frac{E_1}{E_2}\right)^2$
Substitute the known values:
$\frac{n_2}{1000} = \left(\frac{100}{50}\right)^2$
$\frac{n_2}{1000} = (2)^2$
$\frac{n_2}{1000} = 4$
$n_2 = 4 \times 1000 = 4000$
To reduce the margin of error from $100$ to $50$ (i.e., by half), the sample size needs to increase by a factor of $2^2 = 4$. The new sample size required is $4000$.

This problem asks for the probability of drawing three non-defective gears without replacement. The key concept here is conditional probability combined with sampling without replacement, meaning the total number of items and the number of desired items change after each selection.

Understanding the Setup:
We have a total of 10 spur gears.

• Total gears ($N$): 10
• Defective gears ($D$): 1
• Non-defective gears ($N_D$): $10 - 1 = 9$
  We need to draw 3 gears ($n=3$) such that all are non-defective.

Step-by-Step Probability Calculation:
We will calculate the probability for each draw, considering the reduction in the total number of gears and non-defective gears after each successful draw.

1. Probability of the 1st gear being non-defective:
   There are 9 non-defective gears out of 10 total.
   $P(\text{1st is non-defective}) = \frac{9}{10}$

2. Probability of the 2nd gear being non-defective (given the 1st was non-defective):
   After taking one non-defective gear, 8 non-defective gears remain out of a total of 9 gears.
   $P(\text{2nd is non-defective | 1st was non-defective}) = \frac{8}{9}$

3. Probability of the 3rd gear being non-defective (given the first two were non-defective):
   After taking two non-defective gears, 7 non-defective gears remain out of a total of 8 gears.
   $P(\text{3rd is non-defective | 1st \& 2nd were non-defective}) = \frac{7}{8}$

Overall Probability:
To get the probability that all three are non-defective, we multiply these individual probabilities:
$P(\text{All 3 non-defective}) = \frac{9}{10} \times \frac{8}{9} \times \frac{7}{8}$
$P(\text{All 3 non-defective}) = \frac{7}{10}$

Final Result:
Converting the fraction to a decimal and rounding to two decimal places:
$\frac{7}{10} = 0.7$
Rounded to two decimal places, the probability is $0.70$.

Concept: The probability density function of the exponential distribution is, f(t) = λe -λt Where, mean = $\frac{1}{\lambda }$ , t = time, Calculation: Given: Maen = 15 minutes, $\lambda =\frac{1}{15}$ The probability that passenger will miss the flight = P( t >15) P(t > 15) = 1 - P(t ≤ 15) = 1 - $\int_{0}^{15}\lambda e^{-\lambda t}dt$ = 1 -

= $e^{-(\frac{1}{15})15}$ = $\frac{1}{e}$ P( t > 15) = 0.368

Eigenvalues of Matrix A We are given the matrix

. The problem explicitly states that the eigenvalues of matrix $A$ are 5 and 10. Defining Matrix B = A + αI A second matrix, $B$, is defined based on matrix $A$. The relationship is given by $B = A + \alpha I$, where $\alpha$ is a scalar constant, and $I$ represents the $2 \times 2$ identity matrix. The identity matrix $I$ is: $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ Substituting the matrices into the definition of $B$: $B = \begin{bmatrix} 8 & 3 \\ 2 & 7 \end{bmatrix} + \alpha \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ Performing the scalar multiplication and matrix addition: $B = \begin{bmatrix} 8 & 3 \\ 2 & 7 \end{bmatrix} + \begin{bmatrix} \alpha & 0 \\ 0 & \alpha \end{bmatrix} = \begin{bmatrix} 8+\alpha & 3 \\ 2 & 7+\alpha \end{bmatrix}$ Our goal is to find the eigenvalues of this matrix $B$. Eigenvalue Property for Matrix Shifts In linear algebra, there is a useful property concerning the eigenvalues of a matrix modified by adding a scalar multiple of the identity matrix. If $\lambda$ is an eigenvalue of a square matrix $A$, then $\lambda + \alpha$ is an eigenvalue of the matrix $A + \alpha I$. Calculating Eigenvalues for Matrix B We can apply this property directly using the known eigenvalues of matrix $A$. For the eigenvalue $\lambda_1 = 5$ of matrix $A$, the corresponding eigenvalue of matrix $B$ is $5 + \alpha$. For the eigenvalue $\lambda_2 = 10$ of matrix $A$, the corresponding eigenvalue of matrix $B$ is $10 + \alpha$. Therefore, the eigenvalues of matrix $B$ are $5 + \alpha$ and $10 + \alpha$. Mathematical Justification Let's verify this property. Assume $v$ is an eigenvector of matrix $A$ corresponding to the eigenvalue $\lambda$. By definition: $Av = \lambda v$ Now, consider the matrix $B = A + \alpha I$. We apply $B$ to the eigenvector $v$: $Bv = (A + \alpha I)v$ Using the distributive property of matrix multiplication: $Bv = Av + (\alpha I)v$ Since $Iv = v$, we can simplify the second term: $Bv = Av + \alpha v$ Now substitute $Av = \lambda v$: $Bv = \lambda v + \alpha v$ Factor out the common term $v$: $Bv = (\lambda + \alpha) v$ This resulting equation shows that $v$ is also an eigenvector for matrix $B$, and the eigenvalue associated with it is $\lambda + \alpha$. This confirms the property used. Applying this to the given eigenvalues of A (5 and 10), the eigenvalues for B are indeed $5 + \alpha$ and $10 + \alpha$. This matches the second option provided.

Explanation: (3i + 1)x + (4i + 4)y + 5 = 0 3ix + x + 4iy + 4y + 5 = 0 Now, we can write this equation as (x + 4y + 5) + i(3x + 4y) = 0 + 0i On comparing LHS and RHS, We have x + 4y + 5 = 0 ---(1) 3x + 4y +0 = 0 ---(2) Subtracting (1) from (2), We get 2x - 5 = 0 ∴ 2x = 5 x = $\frac{5}{2}$ x = 2.5

Concept: Trapezoidal Rule : When the area under the curve is evaluated, then the total area is divided into small trapezoids instead of rectangles. It is given as $\int_a^b {f(x)} \ dx$ ≈ Tn = \frac {Δ x}{2} [f(x_0) + f(x_n) +2(f(x_1)+f(x_2)+⋯+f(x_{n−1}))] Where, Δx = $\frac {b-a}{n}$ ; Calculation: Given: \mathop \smallint \nolimits_0^{10} y (x) dx = ?; a = 0; b = 10; n = 10; x 0 1 2 3 4 5 6 7 8 9 10 y(x) 5 3 0 -5 -10 -6 0 5 11 18 30 Δx = $\frac {10-0}{10}$ = 1 \mathop \smallint \nolimits_0^{10} y (x) dx = T n T n = \frac {Δ x}{2} [f(x_0) + f(x_n) +2(f(x_1)+f(x_2)+⋯+f(x_{n−1}))] = $\frac {1}{2} [ 5 + 30 +2(3+0-5-10-6+0+5+11+18)]$ = $\frac {1}{2} [ 5 + 30 +2(16)]$ T n = 33.5

Process Control Type-I Error Probability Explained This problem requires us to calculate the combined probability of a Type-I error when multiple independent rules are simultaneously applied to a process control chart. A Type-I error occurs when we incorrectly conclude that a process is out of statistical control when it is actually stable. Understanding Type-I Error in Control Charts In Statistical Process Control (SPC), a Type-I error, often denoted by the Greek letter alpha ($\alpha$), is essentially a false alarm. It means the control chart indicates a process problem (out-of-control state) when no real problem exists. Given Information We have four distinct rules used to detect an out-of-control state. Each rule has its own probability of causing a Type-I error. These probabilities are given as: Rule 1: $P(\text{Type-I Error}_1) = \alpha_1 = 0.005$ Rule 2: $P(\text{Type-I Error}_2) = \alpha_2 = 0.02$ Rule 3: $P(\text{Type-I Error}_3) = \alpha_3 = 0.03$ Rule 4: $P(\text{Type-I Error}_4) = \alpha_4 = 0.05$ Crucially, we are told that these rules are independent of each other. Method for Calculating Combined Probability When multiple rules are used together, the process is considered out-of-control if any one of these rules triggers a signal. Calculating the probability of "at least one" event is often simplified by calculating the probability of the complementary event - that none of the events occur - and subtracting this from 1. In this context, the complementary event is that the process remains in control according to all four rules simultaneously. Let $A_i$ represent the event that Rule $i$ detects an out-of-control state. We are given $P(A_i) = \alpha_i$. The probability that Rule $i$ does not detect an out-of-control state (i.e., it indicates the process is in control) is $P(A_i^c) = 1 - P(A_i) = 1 - \alpha_i$. For Rule 1: $P(A_1^c) = 1 - 0.005 = 0.995$ For Rule 2: $P(A_2^c) = 1 - 0.02 = 0.98$ For Rule 3: $P(A_3^c) = 1 - 0.03 = 0.97$ For Rule 4: $P(A_4^c) = 1 - 0.05 = 0.95$ Since the rules are independent, the probability that the process is considered in control by all rules is the product of these individual probabilities: $P(\text{In Control by All Rules}) = P(A_1^c) \times P(A_2^c) \times P(A_3^c) \times P(A_4^c)$ Step-by-Step Calculation Let's compute the product: $P(\text{In Control by All Rules}) = 0.995 \times 0.98 \times 0.97 \times 0.95$ Performing the multiplication step-by-step: $0.995 \times 0.98 = 0.9751$ $0.9751 \times 0.97 = 0.945847$ $0.945847 \times 0.95 = 0.89855465$ So, the probability that none of the rules signal an out-of-control state is approximately $0.8986$. Final Probability of Overall Type-I Error The overall Type-I error probability, when using all four rules simultaneously, is the probability that at least one rule signals an out-of-control state. This is calculated as: $P(\text{Overall Type-I Error}) = 1 - P(\text{In Control by All Rules})$ $P(\text{Overall Type-I Error}) = 1 - 0.89855465$ $P(\text{Overall Type-I Error}) = 0.10144535$ Rounding this value to three decimal places, we get $0.101$. This represents the total probability of falsely detecting an out-of-control situation when using all four rules concurrently.

Evaluating Retail Store Site Selection Factors This problem requires selecting the most suitable location for a new retail store from four potential sites: A, B, C, and D. The selection is based on a weighted scoring method. Several key factors are considered, each assigned a specific weight reflecting its importance. Each site is then scored out of 100 for every factor. Data for Site Evaluation The table below outlines the factors, their weights, and the scores awarded to each site: Factor Factor Weight Score for site (out of 100) A B C D Average community income 0.4 60 70 80 50 Demand growth potential 0.13 30 80 50 40 Proximity to existing store 0.35 50 10 40 60 Availability of public transport 0.2 40 30 40 20 Calculating Weighted Scores for Each Site To find the best site, we calculate a weighted score for each location. This involves multiplying the score of each factor by its corresponding weight and summing these products. The formula applied is: $\text{Weighted Score} = \sum_{i=1}^{n} (\text{Score}_i \times \text{Weight}_i)$ Here, $\text{Score}_i$ represents the score for the i-th factor, and $\text{Weight}_i$ is the weight assigned to it. Site A Weighted Score Calculation The weighted score for Site A is calculated as follows: $\text{Score}_A = (60 \times 0.4) + (30 \times 0.13) + (50 \times 0.35) + (40 \times 0.2)$ $\text{Score}_A = 24 + 3.9 + 17.5 + 8 = 53.4$ Site B Weighted Score Calculation The weighted score for Site B is calculated as follows: $\text{Score}_B = (70 \times 0.4) + (80 \times 0.13) + (10 \times 0.35) + (30 \times 0.2)$ $\text{Score}_B = 28 + 10.4 + 3.5 + 6 = 47.9$ Site C Weighted Score Calculation The weighted score for Site C is calculated as follows: $\text{Score}_C = (80 \times 0.4) + (50 \times 0.13) + (40 \times 0.35) + (40 \times 0.2)$ $\text{Score}_C = 32 + 6.5 + 14 + 8 = 60.5$ Site D Weighted Score Calculation The weighted score for Site D is calculated as follows: $\text{Score}_D = (50 \times 0.4) + (40 \times 0.13) + (60 \times 0.35) + (20 \times 0.2)$ $\text{Score}_D = 20 + 5.2 + 21 + 4 = 50.2$ Comparison of Site Scores and Final Selection After calculating the weighted scores for all four sites, we compare them to determine the best location: Site A: 53.4 Site B: 47.9 Site C: 60.5 Site D: 50.2 Site C achieved the highest weighted score ($60.5$). This indicates that, based on the provided factors and their weights, Site C offers the most advantageous profile for opening the new retail store compared to sites A, B, and D.

Conept: f(x, y, z) = x 2 + 5y 2 + 5z 2 - 4x + 40y - 40z + 300 For minimum Value, $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = \frac{\partial f}{\partial z} = 0$ Calculation: Given: $f_x = x^2 - 4x$ ∴ $\frac{\partial f}{\partial x}$ = 2x - 4 = 0 2x - 4 =0 So, x = 2 Similarly, for $f_y = 5y^2 + 40y$ ∴ $\frac{\partial f}{\partial y}$ = 10y + 40 = 0 10y + 40 = 0 So, y = -4 Also, for $f_z = 5z^2 - 40z$ ∴ $\frac{\partial f}{\partial z}$ = 10z - 40 = 0 So, z = 4 Now, Putting values of x, y, z in function f(x, y, z), f(x, y, z) = x 2 + 5y 2 + 5z 2 - 4x + 40y - 40z + 300 f(2, -4, 4) = $2^2 + 5(-4)^2 + 5(4)^2 - 4(2) + 40(-4) - 40(4) + 300$ f(2, -4, 4) = 4 + 80 + 80 - 8 - 160 - 160 + 300 f(2, -4, 4) = 136 ∴ the minimum value of function f defined by f(x, y, z) = 136

Concept: $E(x) =\sum_{1}^{\infty }x_iP(x_i)$ Where E(x) is the Expected value, x = random variable, P(x) is probability density function. Calculation: Given: $P(x)=\frac{1}{2^x}$ Expected value, $E(x) =\sum_{1}^{\infty }x_iP(x_i)$ $E(x) =\left ( 1.\frac{1}{2^1} \right )+\left ( 2.\frac{1}{2^2} \right )+\left ( 3.\frac{1}{2^3} \right )+\left ( 4.\frac{1}{2^4} \right )+.....$ < 1 > Dividing both sides by 2, $\frac{E(x)}{2} =\left ( 1.\frac{1}{2^2} \right )+\left ( 2.\frac{1}{2^3} \right )+\left ( 3.\frac{1}{2^4} \right )+\left ( 4.\frac{1}{2^5} \right )+.....$ < 2 > Substracting equation < 2 > from equation < 1 > $E(x)-\frac{E(x)}{2} =\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....$ $\frac{E(x)}{2} = \frac{\frac{1}{2}}{1-\frac{1}{2}}$ $\frac{E(x)}{2} =1$ E(x) = 2

To find the integrating factor for the given differential equation, we first need to recognize its standard form. The given equation is $\dfrac{dy}{dx}+my=e^{-mx}$, which is a first-order linear differential equation in the form $\dfrac{dy}{dx} + P(x)y = Q(x)$.

By comparing, we can identify $P(x) = m$ and $Q(x) = e^{-mx}$. Remember that $m$ is treated as a constant here with respect to $x$.

The formula for the Integrating Factor (IF) for such an equation is $IF = e^{\int P(x) dx}$.

Now, substitute $P(x) = m$ into the formula:
$\int P(x) dx = \int m \, dx$

Since $m$ is a constant, performing the integration gives:
$\int m \, dx = mx$

Therefore, the Integrating Factor (IF) is:
$IF = e^{mx}$

Comparing this result with the given options, we find that option D, $e^{mx}$, is the correct integrating factor.

This is a Cauchy-Euler differential equation of the form $ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0$. Comparing $x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-y=0$ with the general form, we have $a=1$, $b=1$, and $c=-1$.

We assume a solution of the form $y=x^m$. Then, we find the derivatives:
$\frac{dy}{dx} = mx^{m-1}$
$\frac{d^2y}{dx^2} = m(m-1)x^{m-2}$

Substitute these into the differential equation:
$x^2\left(m(m-1)x^{m-2}\right)+x\left(mx^{m-1}\right)-x^m=0$
$m(m-1)x^m+mx^m-x^m=0$
Factor out $x^m$ (assuming $x^m \neq 0$ for a non-trivial solution):
$x^m\left(m(m-1)+m-1\right)=0$

The auxiliary (characteristic) equation is:
$m(m-1)+m-1=0$
$m^{2-}m+m-1=0$
$m^{2-}1=0$

Solving for $m$:
$m^2=1$
$m=\pm\sqrt{1}$
This gives two distinct real roots: $m_1=1$ and $m_2=-1$.

For distinct real roots $m_1$ and $m_2$, the general solution for a Cauchy-Euler equation is given by $y=C_1x^{m_1}+C_2x^{m_2}$.
Substituting the values of $m_1$ and $m_2$:
$y=C_1x^{1+}C_2x^{-1}$
$y=C_1x+\frac{C_2}{x}$

To find $(z_1 + z_2)^2$, we first sum the complex numbers $z_1 = 2 + 3i$ and $z_2 = 4 - 5i$.
$z_1 + z_2 = (2 + 4) + (3 - 5)i = 6 - 2i$
Next, we square this result. Using the binomial expansion $(a - b)^2 = a^2 - 2ab + b^2$, where $a = 6$ and $b = 2i$:
$(6 - 2i)^2 = 6^2 - 2(6)(2i) + (2i)^2$
$(6 - 2i)^2 = 36 - 24i + 4i^2$
Since $i^2 = -1$:
$(6 - 2i)^2 = 36 - 24i + 4(-1)$
$(6 - 2i)^2 = 36 - 4 - 24i$
$(6 - 2i)^2 = 32 - 24i$