Question

find the modulus and argument of (1+i)^2/2+i​

Answer 1

Answer:

Let, z=1−3i1+2i

z=1−3i1+2i×1+3i1+3i

=12+321+3i+2i+6i2

=1+91+5i+6(−1)

=10−5+5i

=2−1+21i

Let z=rcosθ+irsinθ

i.e.,rcosθ=2−1 and rsinθ=21

On squaring and adding, we obtain

r2(cos2θ+sin2θ)=(2−1)2+(21)2