# Archive | ece 493

## solutions to exercises 1.19

Exercise: do the polynomials $$p_0(x)=1,~~p_1(x)=(x-1),~~p_2(x)=(x-1)^2,~~p_3=(x-1)^3$$ form a basis of the vector space of cubic polynomials (with real coefficients)? If so, express $$x^3$$ in this basis.

Solution: use the binomial: $$x^n = \big((x-1)+1\big)^n = \sum_{m=0}^n {n\choose m}(x-1)^m$$ to infer …

## 1.19 Linear spaces

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}} \def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}} \def\eye{\left(\begin{array}{cc}1&0\\0&1\end{array}\right)} \def\bra#1{\langle #1|}\def\ket#1{|#1\rangle}\def\j{\mathbf{j}}\def\dim{\mathrm{dim}} \def\ker{\mathbf{ker}}\def\im{\mathbf{im}}$$

1. Linear (or vector) spaces (over a field $$k$$ – which in our case will be almost always the complex numbers) are sets that satisfy the following standard list of properties:

• There exists

## solutions to exercises 1.17

Exercise 1: describe the curve $$1/z$$, where $$z=1+ti, ~~t\in (-\infty,\infty)$$.

Solution: apply $$\frac{1}{z} = \frac{\bar z}{|z|^2}$$ and get $$\frac{1}{1+ti} = \frac{1-ti}{1+t^2}$$, which is a parametric curve  of the form $$p(t) = x(t)+iy(t)$$ where $$x(t) = \frac{1}{1+t^2},~~~ y(t) = \frac{-t}{1+t^2}$$.…

## 1.17 Complex numbers

$$\def\Real{\mathbb{R}}\def\Comp{\mathbb{C}}\def\Rat{\mathbb{Q}}\def\Field{\mathbb{F}}\def\Fun{\mathbf{Fun}}\def\e{\mathbf{e}}\def\f{\mathbf{f}}\def\bv{\mathbf{v}}\def\i{\mathbf{i}} \def\eye{\left(\begin{array}{cc}1&0\\0&1\end{array}\right)}$$

Complex numbers are expressions $$z=x+\i y$$, where $$\i$$ is the imaginary unit, defined by $$\i^2=-1$$. Note the ambiguity ($$-\i$$ would work just as well).

Complex numbers can be added subtracted, multiplied the usual way, with the …