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Question

Find the modulus and the argument of the complex number z=(1+2i)(13i)

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Solution

We have,
z=(1+2i)(13i)

=(1+2i)(1+3i)(13i)(1+3i)

=1(1+3i)+2i(1+3i)1²(3i)²

=(1+3i+2i+6i²)(1+9)

=(5+5i)10
=12+12i

Now, modulus of the complex number is =12+12i
=(1/2)²+(1/2)²
=1/4+1/4
=12

Now,
tanϕ=Im(z)Re(z)
=∣ ∣ ∣1212∣ ∣ ∣=1

tanϕ=tanπ4
ϕ=π4

ϕ lies on 2nd quadrant so,
arg(z)=πϕ
arg(z)=ππ4
arg(z)=3π4

Hence, this is the answer.

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