Math 0250 Exam 1 Solutions

October 12, 1998

F. Beatrous

- 1.
- (a)
- Express the complex number -2+2
*i*in polar form .

**Solution:**

- (b)
- Find all cube roots of -2+2
*i*. 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

and

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

and

- 2.
- Find complex numbers
*A*and*b*so that the real function is the real part of the complex function*F*(*t*) =*A e*^{bt}.

**Solution:**

*f*(*t*) is the real part of*F*(*t*) =*e*^{t}(1-2*i*)*e*^{it}= (1-2*i*)*e*^{t+it}= (1-2*i*)*e*^{(1+i)t}

- 3.
- Find the inverse of the matrix

Your answer will depend on the parameter*a*.**Solution:**

so

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

**Solution:**

The augmented matrix

row reduces to

which yields the equivalent system

This system can be solved for*x*_{1}and*x*_{3}in terms of*x*_{2}and*x*_{4}. Assigning*x*_{2}and*x*_{4}the arbitrary values*s*and*t*respectively gives

In vector notation, this can be written

A basis for the space of solutions is therefore

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

**Solution:**

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

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

- 5.
- Find all solutions to the differential equation

**Solution:**

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

The integrating factor is

and multiplying through by gives

This can be rewritten

and integration gives

Finally, dividing by*t*^{2}gives