Adding non stoichiometric amount of NH3 to copper solution
Both your conclusions are absolutely correct. 

I'll point out that you'll never get "all" the $\ce{Cu^{2+}}$ to the amine complex. You're trying to get greater than 99% of the $\ce{Cu^{2+}}$ to the amine complex. Because 99% and 99.9% and 99.99% would have virtually the same absorption reading.

To give even more detail, in mildly acid solution the copper will be in a octahedral complex $\ce{Cu(H2O)6^{2+}}$ which is a light blue. As you add ammonia at first $\ce{Cu(OH)2}$ forms, which is then dissolved as the various ammine complexes form. 

$\ce{Cu(H2O)_6^2+ + NH3 &lt;--&gt; Cu(NH3)(H2O)5^2+}$
$\ce{Cu(NH3)(H2O)5^2+ + NH3 &lt;--&gt; Cu(NH3)2(H2O)4^2+}$
$\ce{Cu(NH3)2(H2O)4^2+ + NH3 &lt;--&gt; Cu(NH3)3(H2O)3^2+}$
$\ce{Cu(NH3)3(H2O)3^2+ + NH3 &lt;--&gt; Cu(NH3)4(H2O)2^2+}$

The tetraammine complex, $\ce{Cu(NH3)4(H2O)2^2+}$, is a much darker blue than the corresponding complex with water molecules alone. Thus colorimetry is much more sensitive measuring the tetraammine complex that the water complex. 

The tetraammine complex has shorter Cu-N bonds than the C-O bonds. Thus the four ammine groups are one plane and the two water molecules at the apexes of the square bipyramidal complex. 

Hopefully now the point about not converting all the copper to the tetraammine complex is clear. The various ammine complexes are in equilibrium with each other. Given that $\ce{[NH3] &gt;&gt; [Cu^{2+}]}$  then one can imagine the overall reaction as 

$\ce{Cu(H2O)_6^2+ + 4NH3 &lt;--&gt; Cu(NH3)4(H2O)2^{2+}}$

In extremely concentrated solutions of ammonia the final two ammine groups will attach. 

$\ce{Cu(NH3)4(H2O)2^{2+} + NH3 &lt;--&gt; Cu(NH3)5(H2O)^2+}$
$\ce{Cu(NH3)5(H2O)^{2+} + NH3 &lt;--&gt; Cu(NH3)6^2+}$