Solution Notes -- Example Problem Set 1

1.

The real part of z₃ is ac-bd. Its imaginary part is ad+bc.

2.

Modulus $\Vert{z}_{1}\Vert = \sqrt{a^{2}+b^{2}}$.

Modulus $\Vert{z}_{3}\Vert = \sqrt{(ac-bd)^{2}+(ad+bc)^{2}}$ , or equivalently, $\Vert{z}_{3}\Vert = \sqrt{(a^{2}+b^{2})(c^{2}+d^{2})}$.

3.

Angle $\angle {z}_{2} = \tan^{-1}(\frac{d}{c})$.

4.

In complex polar form, ${z}_{1} = \Vert{z}_{1}\Vert e^{i \angle {z}_{1}}$.

5.

As is clear from describing these variables in complex polar form,

$${z}_{3} = \Vert{z}_{1}\Vert e^{i\angle{z}_{1}} \Vert{z}_{2}\V... ...Vert \Vert{z}_{2} \Vert e^{i (\angle {z}_{1}+\angle {z}_{2})},$$

and so the angles of the two complex variables just add. This is a rotation in the complex plane. Since the two moduli both happen to equal 1, the modulus of the product z₁z₂ is also $\Vert{z}_{3}\Vert = 1$.

6.

If ${z} = e^{2\pi i/5}$ is multiplied by itself five times, its angle is just added to itself five times, producing $e^{2\pi i} = 1$.

7.

The real part of $f(x) = e^{2\pi i \omega x}$ is the function $\cos(2\pi \omega x)$. Its imaginary part is the function $\sin(2\pi \omega x)$.

8.

If the complex exponential $f(x) = e^{2\pi i \omega x}$ is operated upon by any linear operator, its functional form cannot change. The most dramatic thing that can happen to it is that it gets multiplied by a complex constant. This means that only its amplitude and phase can be affected. Complex exponentials are the eigenfunctions of linear systems.