We start with $$\frac{(-15j+15)(12Mj+16M)}{9M+16M}$$ where M standing for $10^6$, and $j=\sqrt{-1}$.
After multiplying, I get $$\frac{180M-240Mj+180Mj+240M}{25M}$$ Then, $$\frac{420M-60Mj}{25M}$$. This is in the fourth quadrand. The vector is magnitude is $17M$. And the angle is $-8.13^{\circ}$, so $17M\angle-8.13^{\circ}$, in polar notation.
But if I converted all the rectangular notations into polar first, I would get something different. I can't spot my arithmetic error.
$$-15j+15=15 \angle {-45^{\circ}}$$ $$12Mj+16M=20M \angle {53^{\circ}}$$ $$25M=25\angle0^{\circ}$$
So put them all together:
$$\frac{(15 \angle {-45^{\circ}})(20M \angle {53^{\circ}})}{25\angle0^{\circ}}$$
To multiply in polar, I multipled the magnitudes together, and added the angles. To divide, I divide the top magnitude by the bottom magnitude, and substract the bottom angle from the top angle. But I got $12M\angle8^{\circ}$
$12Mj + 16M = 20M\angle 36^\circ$, looks like you mixed up $y$ and $x$ for arctan. Also, $-15j + 15 = (15\sqrt{2})\angle-45^\circ$, not $15\angle-45^\circ$. I think those are the only two mistakes.