If Z=∣∣
∣∣25−i7+i5+i23−i7−i3+i7∣∣
∣∣ and arg (z) = θ then θ =
Z=∣∣ ∣∣25−i7+i5+i23−i7−i3+i7∣∣ ∣∣
Now we can solve this determinant either using Sarrus rule or using Cofactors.
Let’s do it using cofactors
So Z=2((2×7)−(3+i)(3−i))↓ Cofactor for a11=2+(5−i)×(−1)1+2((5+i)×7−(3−i)(7−i))↓ Cofactor for a12=5−i+(7+i)(−1)1+3((5+i)(3+i)−2(7−i))↓ Cofactor for a13=7+i
+(5−i)x(−1)(35+7i−(21−3i7i1))+(7+i)(1)(15+8i)114+2i)=−94
Which is a negative real number and we know that for a number lying on negative real axis, the argument is π, which is correct answer.