1.6. Exercises

You should try the following exercise questions first, then check with the answers.

For detailed solutions, you can download the full solutions for Chapter 2 Exercises.

Exercise 1.1

Simplify the following expressions used to calculate \(z\), and find the real and imaginary components of \(z\):

1. \(\displaystyle z=\frac{4+3\i}{3-4\i}\),

2. \(\displaystyle z=e^{\i\theta}-e^{-\i\theta}\),

3. \(\displaystyle z=e^{\i 2\theta} \cdot e^{-\i\theta}\),

4. \(\displaystyle z=\frac{2e^{\i\theta}(e^{\i\theta}-1)}{3e^{\i\theta}-1}\).

Solution to Exercise 1.1

1. \(z=\i\)

2. \(z=\i 2\sin\theta\)

3. \(z=\cos\theta+\i\sin\theta\)

4. \(z=\dfrac{4\cos\theta-\cos 2\theta-3}{5-3\cos\theta}+\i\dfrac{4\sin\theta-\sin 2\theta}{5-3\cos\theta}\)

Exercise 1.2

For the following second-order homogeneous linear difference equation

\[y_{n+2} - 2 a y_{n+1}+ a^2 y_n =0, \quad a\in \R,~ a\ne0,\]

prove that

\[y_n = c_1 a^n + c_2 n a^n\]

is the solution to the equation.

Solution to Exercise 1.2

According to the question, the members of the sequence at \(n\), \(n+1\) and \(n+2\) are:

• \(y_n = c_1 a^n + c_2 n a^n\)

• \(y_{n+1} = c_1 a^{n+1} + c_2 (n+1) a^{n+1}\)

• \(y_{n+2} = c_1 a^{n+2} + c_2 (n+2) a^{n+2}\)

Substitute the above three members into the left hand side (LHS) of the equation, so

\[\begin{split}\displaystyle \begin{aligned} \text{LHS} & = y_{n+2} - 2 a y_{n+1}+ a^2 y_n \\ & = \left(c_1 a^{n+2} + c_2 (n+2) a^{n+2} \right) -2a \left(c_1 a^{n+1} + c_2 (n+1) a^{n+1} \right) + a^2 \left(c_1 a^n + c_2 n a^n \right) \\ & = c_1 a^{n+2} + c_2 (n+2) a^{n+2} -2 c_1 a^{n+2} -2 c_2 (n+1) a^{n+2} + c_1 a^{n+2} + c_2 n a^{n+2} \\ & = a^{n+2} \left[c_1+c_2 (n+2) - 2 c_1 - 2 c_2 (n+1) + c_1 + c_2 n\right] \\ & = a^{n+2} \left\{ c_1 (1-2+1) + c_2 \left[(n+2)-2(n+1)+n\right]\right\} \\ & = 0 \end{aligned} \end{split}\]

Q.E.D.

Exercise 1.3

For each of the following difference equations, form the characteristic equation, and from its roots determine the corresponding general solution:

1. \(y_{j+1}=y_{j-1}\)

2. \(y_{j+1}=4y_j-3y_{j-1}\)

3. \(y_{j+1}=2y_{j-1}-y_j\)

4. \(y_{j+1}+9y_j - 9y_{j-1}-y_{j-2}=0\)

Solution to Exercise 1.3

1. \(y_j = c_1 + c_2 (-1)^j\)

2. \(y_j = c_1 + c_2 (3)^j\)

3. \(y_j = c_1 + c_2 (-2)^j\)

4. \(y_j = c_1 + c_2 \left(-5+2\sqrt{6}\right)^j+c_3\left(-5-2\sqrt{6}\right)^j\)

Exercise 1.4

For each of the following difference equations, form the characteristic equation, and from its roots determine the corresponding general solution:

1. \(y_{n+2}-9y_{n+1}+20y_n=0\)

2. \(y_{n+2}+y_{n}+y_{n-1}=0\)

3. \(y_{n+2}=\dfrac{y_{n+1}+y_{n-1}}{2}\)

4. \(y_{n+2}=-\dfrac{y_{n+1}+y_{n-1}}{2}\)

5. \(y_{n+4}-16 y_n=0\)

6. \(y_{n+4}+16 y_n=0\)

Solution to Exercise 1.4

1. \(y_n=c_1 (4)^n + c_2 (5)^n\)

2. \(y_n=c_1 \left(\dfrac{-1+\sqrt{3}\i}{2}\right)^n + c_2 \left(\dfrac{-1-\sqrt{3}\i}{2}\right)^n\)

3. \(y_n=c_1+c_2 \left(\dfrac{-1+\sqrt{7}\i}{4}\right)^n + c_3 \left(\dfrac{-1-\sqrt{7}\i}{4}\right)^n\)

4. \(y_n=c_1 (-1)^n+c_2 \left(\dfrac{1+\sqrt{7}\i}{4}\right)^n + c_3 \left(\dfrac{1-\sqrt{7}\i}{4}\right)^n\)

5. \(y_n = c_1 (2)^n + c_2 (-2)^n + 2^n \left(c_3 \cos\dfrac{n\pi}{2}+c_4 \sin\dfrac{n\pi}{2}\right)\)

6. \(y_n=2^n\left(c_1\cos\dfrac{n\pi}{4}+c_2\sin\dfrac{n\pi}{4}\right)+2^n\left(c_3\cos\dfrac{3n\pi}{4}+c_4\sin\dfrac{3n\pi}{4}\right)\)

Exercise 1.5

Find the general solution to the difference equation

\[y_{n}=y_{n-1} + y_{n-2},\]

where \(y_0=0\) and \(y_1=1\).

Solution to Exercise 1.5

\(\displaystyle y_n=\frac{1}{\sqrt{5}}\left[ \left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n \right]\)