Question

convert each of the numbers into polar form:
a)1-i
b)root 3 +i
PLEASE

Answer 1

We know that if Z = (x + iy) is a non-zero complex number then:

Z = (rcosθ + irsinθ) is called the Polar form of the complex number (x + iy).

where x = rcosθ and y = rsinθ and r = square root of x² + y² and θ is the Principal argument which lies in the interval - π ≤ θ ≤ π

a) given the complex number  1 - i

comparing the above complex number with x + iy we can notice that x = 1 and y = -1 and we know that r = square root of x² + y²

⇒ r = square root of (1 + 1) = √2

we know that x = rcosθ

⇒ 1 = rcosθ but we got r = √2

⇒ cosθ = 1/√2

we know that y = rsinθ

⇒ -1 = rsinθ but we got r = √2

⇒ sinθ = -1/√2

as θ is the principal argument - it lies between the interval -π ≤ θ ≤ π

the value of θ for which sinθ = -1/√2 and cosθ = 1/√2 in the range of - π ≤ θ ≤ π is -π/4 because sin(-π/4) = -1/√2 and cos(-π/4) = 1/√2

substituting the all the calculated values in Polar form (rcosθ + rsinθ) we get :

= √2cos(-π/4) + i √2sin(-π/4)

So the Polar form of (1 - i) is √2cos(-π/4) + i√2sin(-π/4))

b) given the complex number  √3 + i

comparing the above complex number with x + iy we can notice that x = √3 and y = 1 and we know that r = square root of x² + y²

⇒ r = square root of (3 + 1) = 2

we know that x = rcosθ

⇒ √3 = rcosθ but we got r = 2

⇒ cosθ = √3/2

we know that y = rsinθ

⇒ 1 = rsinθ but we got r = 2

⇒ sinθ = 1/2

as θ is the principal argument - it lies between the interval -π ≤ θ ≤ π

the value of θ for which sinθ = 1/2 and cosθ = √3/2 in the range of - π ≤ θ ≤ π is π/6 because sin(π/6) = 1/2 and cos(π/6) = √3/2

substituting the all the calculated values in Polar form (rcosθ + irsinθ) we get :

= 2cos(π/6) + i 2sin(π/6)

So the Polar form of (√3 + i) is 2cos(π/6) + i 2sin(π/6)

Answer 2

follow the correct answer