$2$'s, $10$'s and $16$'s complement

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I want to find the $2$'s, $10$'s and $16$'s complement of numbers: $-63, -83$ and $-127$. I know how to find $2$'s complement. What confuses me is $10$'s and $16$'s complement. Can anyone please verify my computation?

So, I first compute the $2$'s complement as:

$-63 = (1000001)_{2c}$

$-83 = (10101101)_{2c}$

$-127 = (100000001)_{2c}$

Now, the $16$'s complement is computed by converting the above obtained 2's complement binary to hex as:

$-63 = (41)_{16c}$

$-83 = (AD)_{16c}$

$-127 = (81)_{16c}$

Now, the $10$'s complement is computed as

$-63 = 10^2-63=(37)_{10c}$

$-83 = 10^2-83 = (17)_{10c}$

$-127 = 10^3-127 = (873)_{10c}$