The problem has two parts. The first part asks to determine the modulus and argument of the complex numbers $z_1 = 1 - i\sqrt{3}$ and $z_2 = 2 - 2i$. The second part involves solving the differential equation $y'' + \frac{1}{4}y = y'$ and finding a function $f$ that is a solution to this differential equation, given that its curve passes through the point $A(0,4)$ and the tangent to the curve at $x=2$ is parallel to the x-axis.

1. Problem Description

The problem has two parts. The first part asks to determine the modulus and argument of the complex numbers $z_1 = 1 - i\sqrt{3}$ and $z_2 = 2 - 2i$. The second part involves solving the differential equation $y'' + \frac{1}{4}y = y'$ and finding a function $f$ that is a solution to this differential equation, given that its curve passes through the point $A(0,4)$ and the tangent to the curve at $x=2$ is parallel to the x-axis.

2. Solution Steps

Part 1: Complex Numbers

For $z_1 = 1 - i\sqrt{3}$:

The modulus of $z_1$ is given by $|z_1| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$.

To find the argument of $z_1$, we have $\tan(\theta) = \frac{-\sqrt{3}}{1} = -\sqrt{3}$. Since the real part is positive and the imaginary part is negative, $z_1$ lies in the fourth quadrant. Thus, $\theta = -\frac{\pi}{3}$.

For $z_2 = 2 - 2i$:

The modulus of $z_2$ is given by $|z_2| = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

To find the argument of $z_2$, we have $\tan(\theta) = \frac{-2}{2} = -1$. Since the real part is positive and the imaginary part is negative, $z_2$ lies in the fourth quadrant. Thus, $\theta = -\frac{\pi}{4}$.

Part 2: Differential Equation

The given differential equation is $y'' + \frac{1}{4}y = y'$, which can be rewritten as $y'' - y' + \frac{1}{4}y = 0$.

The characteristic equation is $r^2 - r + \frac{1}{4} = 0$.

We can solve this equation using the quadratic formula or by recognizing that it is a perfect square: $(r - \frac{1}{2})^2 = 0$. So, $r = \frac{1}{2}$ is a repeated root.

The general solution is $y(x) = c_1e^{\frac{1}{2}x} + c_2xe^{\frac{1}{2}x}$, where $c_1$ and $c_2$ are constants.

We are given that the curve passes through $A(0,4)$, so $y(0) = 4$.

$4 = c_1e^0 + c_2(0)e^0 = c_1$. Thus, $c_1 = 4$.

So, $y(x) = 4e^{\frac{1}{2}x} + c_2xe^{\frac{1}{2}x}$.

The tangent to the curve at $x=2$ is parallel to the x-axis, which means $y'(2) = 0$.

First, we find $y'(x)$:

$y'(x) = 2e^{\frac{1}{2}x} + c_2e^{\frac{1}{2}x} + \frac{1}{2}c_2xe^{\frac{1}{2}x} = (2+c_2)e^{\frac{1}{2}x} + \frac{1}{2}c_2xe^{\frac{1}{2}x}$

Now, we plug in $x=2$ and set $y'(2) = 0$:

$0 = (2+c_2)e^1 + \frac{1}{2}c_2(2)e^1 = (2+c_2)e + c_2e = (2+2c_2)e$

$2+2c_2 = 0$, so $2c_2 = -2$, which gives $c_2 = -1$.

Thus, the function is $f(x) = y(x) = 4e^{\frac{1}{2}x} - xe^{\frac{1}{2}x} = (4-x)e^{\frac{1}{2}x}$.

3. Final Answer

$|z_1| = 2$, $\arg(z_1) = -\frac{\pi}{3}$.

$|z_2| = 2\sqrt{2}$, $\arg(z_2) = -\frac{\pi}{4}$.

$f(x) = (4-x)e^{\frac{1}{2}x}$.