Write the exponential form and the principal argument of the following complex numbers.(ii) z = (√3 − i)^6

To find the exponential form and the principal argument of the complex number z = (√3 - i)^6, we first need to express the complex number in polar form.

Step 1: Convert to Polar Form The polar form of a complex number is z = r(cos θ + i sin θ), where r is the magnitude of z and θ is the argument of z.

The magnitude r is given by √(a² + b²), where a and b are the real and imaginary parts of the complex number, respectively. In this case, a = √3 and b = -1, so r = √((√3)² + (-1)²) = 2.

The argument θ is given by arctan(b/a). In this case, θ = arctan((-1)/(√3)) = -π/6.

So, the polar form of the complex number √3 - i is z = 2(cos(-π/6) + i sin(-π/6)).

Step 2: Raise to the Power of 6 Using De Moivre's theorem, (r(cos θ + i sin θ))^n = r^n (cos(nθ) + i sin(nθ)).

So, z^6 = 2^6 (cos(6*(-π/6)) + i sin(6*(-π/6))) = 64 (cos(-π) + i sin(-π)).

Step 3: Convert to Exponential Form The exponential form of a complex number is z = re^(iθ), where r is the magnitude of z and θ is the argument of z.

So, the exponential form of z^6 is z^6 = 64e^(i(-π)).

Step 4: Find the Principal Argument The principal argument is the value of θ in the interval (-π, π]. In this case, the principal argument of z^6 is -π.

So, the exponential form of the complex number z = (√3 - i)^6 is z^6 = 64e^(i(-π)), and the principal argument is -π.