I've never had to typeset complex numbers before and I'm finding that I'm uncertain about what best-practices are. My question is really about typesetting just i. (Hence my title referring to imaginary as opposed to complex numbers.)
I would like to be consistent with the textbook which uses a slantstyle. But aside from the choices of the textbook, I'm curious about what others think: Should it be upright? Should it be italic like a variable name?
When I use the default math style, I don't like the appearance, particularly in conjunction with exponents. It looks too crowded and busy to me. Is this just because I'm not used to typesetting for complex numbers? Or, is there some kind of italic correction I could do that would fix things: \/
seems to be ignored in math mode.
Here's my minimal working example:
\documentclass{article}
\usepackage{amsmath}
\pagestyle{empty}
\usepackage[margin=2.25in]{geometry}
\setlength{\parindent}{0pt}
%%
\newcommand{\mi}{\mathrm{i}} %% roman "i"
\newcommand{\di}{i} %% default math "i"
\begin{document}
\verb=\mathrm= style: (not consistent with font choice of textbook)
\begin{align*}
\mi^0 &= 1 \\
\mi^1 &= \mi \\
\mi^2 &= -1 \\
\mi^3 &= -\mi
\end{align*}
Default math style: (better matches the style of the textbook, but already looking crowded.)
\renewcommand{\di}{i}
\begin{align*}
\di^0 &= 1 \\
\di^1 &= \di \\
\di^2 &= -1 \\
\di^3 &= -\di
\end{align*}
Whichever choice, the following looks too busy.
\begin{align*}
\mi^n &= \mi^{4\times k + r} = \mi^{4\times k} \times \mi^4 = (\mi^4)^k \times \mi^r = 1^k \mi^r = \mi^r \\
\di^n &= \di^{4\times k + r} = \di^{4\times k} \times \di^4 = (\di^4)^k \times \di^r = 1^k \di^r = \di^r
\end{align*}
And if I change the \verb=\times= to \verb=\cdot= it looks even worse:
\[
\di^n = \di^{4\cdot k + r} = \di^{4\cdot k} \cdot \di^4 = (\di^4)^k \cdot \di^r = 1^k \di^r = \di^r
\]
\end{document}
I know I could completely drop using \times
or \cdot
but for my particular audience I want to emphasize the multiplication.
I think it's the dot on the $i$
my eye is visually objecting to (in which case there's not much to do about it, I guess).
\newcommand{\di}{{i\mkern1mu}}
to give some room. – egreg Dec 8 '12 at 16:48i
you can use\imath
, to get more freedom. – Manuel Dec 8 '12 at 16:50{i\mkern1mu}
; otherwisei\mkern1mu^2
would have the exponent to an empty subformula. – egreg Dec 8 '12 at 17:03\def\I{i}
and simply use\I
everywhere in your math. You can fix your one macro later, after finishing your manuscript. After many similar problems, I found this helps me maintain productivity. – Alex Nelson Dec 8 '12 at 20:33