Given: z = (1 + 2i)(i – 1)

Firstly, we calculate the (1 + 2i)(i – 1) and then find the modulus

So, we open the brackets,

1(i – 1) + 2i(i – 1)

= 1(i) + (1)(-1) + 2i(i) + 2i(-1)

= i – 1 + 2i² – 2i

= - i – 1 + 2(-1) [∵ i² = - 1]

= - i – 1 – 2

= - i – 3

Now, we have to find the modulus of (-3 - i)

So,