Answer :

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,


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