modulus of 1+2i/1-3i
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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
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