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)
