powershell variable syntax $($a)?

$() is a subexpression operator. It means "evaluate this first, and do it separately as an independent statement".

Most often, its used when you're using an inline string. Say:

$x = Get-ChildItem C:\;
$x | ForEach-Object {
    Write-Output "The file is $($_.FullName)";
}

Compare that to:

$x = Get-ChildItem C:\;
$x | ForEach-Object {
    Write-Output "The file is $_.FullName";
}

You can also do things like $($x + $y).ToString(), or $(Get-Date).AddDays(10).

Here, without the subexpression, you'd get $a:\calendar. Well, the problem there is that the colon after a variable is an operator. Specifically, the scope operator. To keep PowerShell from thinking you're trying to look for a variable in the a namespace, the author put the variable in a subexpression.

As far as I've been able to tell using PS for the past few years, parentheses without the dollar sign are also essentially subexpressions. They won't be evaluated as a subexpression when within a string, but otherwise they usually will. It's kind of a frustrating quirk that there's no clear difference.

The $() is the subexpression operator. It causes the contained expressions to be evaluated and it returns all expressions as an array (if there is more than one) or as a scalar (single value). Plain parentheses () are simply the mathematical precedence operator and therefore just work in arithmetic expressions to give precedence

Notably, $() is interpreted within a string, whereas () will be just be taken literally - it has no special meaning. So the following:

$a = "Hello"
"$($a.Length)"

gives

5

whereas

"($a.Length)"

gives

"(Hello.Length)"

As I said, the $() can consist of multiple expressions, all of which are returned as output. () does not do this. So this is an error as the contents are not an arithmetic expression:

(1;2)

whereas this

$(1;2)

evaluates to an array and outputs:

1
2

The expression $($a) is evaluated before the application of the trailing scope : operator and prevents the scope being applied directly to a, instead $a is evaluated first and then the scope is applied.