Let z=x+iy Given, ‌|z+3−i|=1 ‌⇒‌‌|(x+3)+(y−1)i|=1 ‌⇒‌‌√(x+3)2+(y−1)2=1 ‌⇒‌‌(x+3)2+(y−1)2=1 Above is the equation of circle with centre ≡(−3,1) and radius =1. Also, arg(z)=π⇒tan−1|‌
y
x
|=π ⇒‌‌|‌
y
x
|=tan‌π=0⇒y=0
By Eq. (i), ⇒x‌=−3 ∴‌z‌=−3+0i ⇒‌ Modulus of ‌z‌=3