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 ((2+3i)^(2))/((1-i)^(2))=(-5+12i)/(-2i)\left((i)/(i) \right)=(-5i+12i^2)/(-2i^2)\\\\=(-12-5i)/(2)=-6-2.5i`

The modulus is

`r=√((-6)^2+(-2.5)^2)=6.5`

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

`\theta=\pi-\tan^(-1) \left((-2.5)/(-6) \right)=\pi-\tan^(-1) \left((5)/(12) \right)`

So, the answer is:

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