Math 0250 Exam 1 Solutions

October 12, 1998

F. Beatrous

1.
(a)
Express the complex number -2+2i in polar form $re^{i\theta}$.


Solution:
$-2+2i=\sqrt8 e^{i3\pi/4}=2\sqrt2 e^{i3\pi/4}$


(b)
Find all cube roots of -2+2i. You may leave your answer in polar form if you wish, but you should include each cube root exactly once in your list of answers.


Solution:
Write

\begin{displaymath}-2+2i =2\sqrt2 e^{i(\frac{3\pi}4+2\pi k)}
\end{displaymath}

and

\begin{displaymath}(-2+2i)^{1/3} = (2\sqrt2)^{1/3} e^{i(\frac{3\pi}4+2\pi k)\frac13}
= \sqrt2 e^{i(\frac\pi4+\frac{2\pi k}3)}
\end{displaymath}

with k=0, 1, 2. In Cartesian form, these three values are

\begin{displaymath}\sqrt2 e^{i\frac\pi4} = 1+i
\end{displaymath}


\begin{displaymath}(1+i)e^{i\frac{2\pi}3} = (1+i)\left(\frac12+\frac{\sqrt3}2 i\right)
= \frac{1-\sqrt3}2 + \frac{1+\sqrt3}2 i
\end{displaymath}

and

\begin{displaymath}(1+i) e^{i\frac{4\pi}3} = (1+i)\left(\frac12-\frac{\sqrt3}2 i\right)
= \frac{1+\sqrt3}2 + \frac{1-\sqrt3}2 i
\end{displaymath}

2.
Find complex numbers A and b so that the real function $f(t)=e^t(\cos t + 2 \sin t)$ is the real part of the complex function F(t) = A ebt.


Solution:
f(t) is the real part of

F(t) = et (1-2i)eit = (1-2i) et+it = (1-2i) e(1+i)t

3.
Find the inverse of the matrix

\begin{displaymath}A = \left[\begin{array}{rrr}1&0&0\\ 0&1&a\\ 1&0&1\end{array}\right].
\end{displaymath}

Your answer will depend on the parameter a.

Solution:

\begin{displaymath}\left[\begin{array}{rrr}1&0&0\\ 0&1&a\\ 1&0&1\end{array}\righ...
...ft.\begin{array}{rrr}1&0&0\\ a&1&-a\\ -1&0&1\end{array}\right]
\end{displaymath}

so

\begin{displaymath}A^{-1} = \left[\begin{array}{rrr}1&0&0\\ a&1&-a\\ -1&0&1\end{array}\right]
\end{displaymath}

4.
(a)
 Find a basis for the solution space to the system

\begin{displaymath}\left[\begin{array}{rrrr}2&4&1&3\\ 1&2&0&1\end{array}\right]
...
...ray}\right]
= \left[\begin{array}{r}0\\ 0\end{array}\right].
\end{displaymath}


Solution:
The augmented matrix

\begin{displaymath}\left[\begin{array}{rrrr}2&4&1&3\\ 1&2&0&1\end{array}\right\vert
\left.\begin{array}{r}0\\ 0\end{array}\right]
\end{displaymath}

row reduces to

\begin{displaymath}\left[\begin{array}{rrrr}1&2&0&1\\ 0&0&1&1\end{array}\right\vert
\left.\begin{array}{r}0\\ 0\end{array}\right]
\end{displaymath}

which yields the equivalent system

\begin{eqnarray*}x_1+2x_2+x_4 &=& 0\\
x_3+x_4 &=& 0
\end{eqnarray*}


This system can be solved for x1 and x3 in terms of x2 and x4. Assigning x2 and x4 the arbitrary values s and t respectively gives

\begin{eqnarray*}x_1 &=& -2 x_2 - x_4 = -2s - t\\
x_2 &=& s\\
x_3 &=& -x_4 = -t\\
x_4 &=& -t
\end{eqnarray*}


In vector notation, this can be written

\begin{displaymath}X = \left[\begin{array}{c}x_1\\ x_2\\ x_3\\ x_4\end{array}\ri...
...]
+ t \left[\begin{array}{r}-1\\ 0\\ -1\\ 1\end{array}\right]
\end{displaymath}

A basis for the space of solutions is therefore

\begin{displaymath}\left\{\left[\begin{array}{r}-2\\ 1\\ 0\\ 0\end{array}\right]...
...eft[\begin{array}{r}-1\\ 0\\ -1\\ 1\end{array}\right] \right\}
\end{displaymath}

(b)
Find all solutions (if there are any) to the system

\begin{displaymath}\left[\begin{array}{rrrr}2&4&1&3\\ 1&2&0&1\end{array}\right]
...
...ray}\right]
= \left[\begin{array}{r}1\\ 0\end{array}\right].
\end{displaymath}


Solution:
You can come up with a single solution by trial and error. This one will do:

\begin{displaymath}X_p = \left[\begin{array}{r}0\\ 0\\ 1\\ 0\end{array}\right]
\end{displaymath}

Combining this with the general solution to the associated homogeneous system found in part 4a give the general solution

\begin{displaymath}X = X_p + X_h = \left[\begin{array}{r}0\\ 0\\ 1\\ 0\end{array...
...
+ t \left[\begin{array}{r}-1\\ 0\\ -1\\ 1\end{array}\right].
\end{displaymath}

5.
Find all solutions to the differential equation

\begin{displaymath}t \frac{du}{dt} + 2 u = t.
\end{displaymath}


Solution:
First, divide through by t to put the equation in standard form:

\begin{displaymath}\frac{du}{dt} + \frac2t u = 1.
\end{displaymath}

The integrating factor is

\begin{displaymath}\mu = e^{\int \frac2t\,dt} = e^{2\log t} = t^2
\end{displaymath}

and multiplying through by $\mu$ gives

\begin{displaymath}t^2 \frac{du}{dt} + 2t u = t^2.
\end{displaymath}

This can be rewritten

\begin{displaymath}\frac{d}{dt} (t^2 u) = t^2
\end{displaymath}

and integration gives

\begin{displaymath}t^2 u = \frac{t^3}3 + C.
\end{displaymath}

Finally, dividing by t2 gives

\begin{displaymath}u = \frac t3 + \frac C{t^2}.
\end{displaymath}



Frank Beatrous
1998-10-13