# What is the absolute value of the number (1+i)(1+2i)(1+3i)?

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### 2 Answers

First find the product of (1 + i)(1 + 2i)(1 + 3i)

z = (1 + 3i - 2)(1 + 3i)

=> z = (-1 + 3i)(1 + 3i)

=> z = 9i^2 - 1

=> z = -9 - 1

=> z = -10

**The absolute value of -10 is 10**

We'll multiply the first two factors:

(1+i)*(1+2i) = 1 + i + 2i + 2i^2

Since i^2 = -1, we'll have:

(1+i)*(1+2i) = 1 + 3i - 2

(1+i)*(1+2i) = -1 + 3i

Now, we'll multiply both sides by (-1+3i):

(1+i)*(1+2i)*(1+3i) = (-1 + 3i)*(1 + 3i)

(1+i)*(1+2i)*(1+3i) = (3i)^2 - 1^2

(1+i)*(1+2i)*(1+3i) = -9 - 1 = -10

**The absolute value of the number (1+i)*(1+2i)*(1+3i) = |-10| = 10.**