Question about delayed sampled sinusoid math expression

Question

I have been studying the digital audio processing by using the book <Designing Audio Effect Plugins in C++>.

For analog Sinusoid:

Complex Sinusoid = $e^{jωt}$

Delayed Sinusoid = $e^{jω(t−n)} = e^{jwt} * e^{-jwn}$, a delay of n seconds

For digital sampled version:

sampled complex sinusoid = $e^{jωnT}$, T is interval for each sample, n is the index of sample

I understand all above, but I got confused about the delayed sampled sinusoid which described as: $e^{jω ( nT −M )}$, M = samples of delay

But I think it should be described as $e^{jωT( n − M )}$, since the T is a constant for a fixed sample rate, n and M has the same unit.

At first I thought that maybe a typo, but the following computation parts of the book are all using the $e^{jω ( nT −M )}$ as basis.

Anyone can explain it for me?

Answer 1

Indeed a typo

$e^{j \omega ((n - k)T)}$ would be the delay for $k$ samples, the $M$ in the expression $e^{j \omega(nT - M)}$ is the delay (in seconds if $\omega$ is in $rad/s$)