Let Z = (1+3i)/(1-2i) .Then the argument of z lies in

Solution

To find the argument of Z = (1+3i)/(1-2i), we first need to express Z in the form a+bi, where a and b are real numbers. We multiply the numerator and the denominator by the conjugate of the denominator Z = [(1 + 3i)(1 + 2i)] / [(1 – 2i)(1 + 2i)] Denominator:
(1 – 2i)(1 + 2i) = 1² – (2i)² = 1 – (–4) = 1 + 4 = 5 Numerator:
(1 + 3i)(1 + 2i) =
= 1×1 + 1×2i + 3i×1 + 3i×2i
= 1 + 2i + 3i + 6i²
= 1 + 5i + 6(–1)
= 1 + 5i – 6 = –5 + 5i So,
Z = (–5 + 5i) / 5 = –1 + i Now we have Z in the form a+bi, where a=−1 and b=1. The argument of a complex number z=a+bi is the angle θ such that a=rcosθ and b=rsinθ, where r= ∣ z ∣ = √(a² + b²) In our case, a=−1 and b=1. Since the real part is negative (a=−1<0) and the imaginary part is positive (b=1>0), the complex number Z lies in the second quadrant of the complex plane. To find the argument θ, we can use tan θ = b/a = 1/(-1) = -1 The angles whose tangent is −1 are 3π/4 + nπ , where n is an integer. Since the complex number is in the second quadrant, the argument θ must lie in the interval (π/2, π] For n=0, θ=3π/4​, which lies in the second quadrant. Therefore, the argument of Z lies in the second quadrant.