Question

Please help me do this question

Answer 1

$(2+3i)^2 =4+12i+9i^2=4+12i-9=-5+12i\\\\(1-i)^2=1-2i+i^2 = 1-2i-1=-2i$

$\therefore \frac{(2+3i)^{2}}{(1-i)^{2}}=\frac{-5+12i}{-2i}\left(\frac{i}{i} \right)=\frac{-5i+12i^2}{-2i^2}\\\\=\frac{-12-5i}{2}=-6-2.5i$

The modulus is

$r=\sqrt{(-6)^2+(-2.5)^2}=6.5$

Since the complex number lies in the third quadrant, the argument is

$\theta=\pi-\tan^{-1} \left(\frac{-2.5}{-6} \right)=\pi-\tan^{-1} \left(\frac{5}{12} \right)$

So, the answer is:

$\boxed{6.5e^{i\left(\pi-\tan^{-1} \left(\frac{5}{12} \right) \right)}}$