Polar and Euler form

Question

Q1)

 

Q2)

Answer 1

Q1)

a) The complex number w = 1 - i

The polar form of complex number z = x + iy is z = a [cos(b) + i sin(b)]

In this case x = 1, y = - 1

a = | w | = √ (x²  + y²)

a = √ (1²  + (-1)²)

a = √2

Since x > 0, use the formula b = tan^-1(y/x)

b = tan^-1 (-1/1)

b = tan^-1 (-1)

b = - 45

b = - π/4

Polar form of w = 1 - i is √2 [cos (- π/4) + i sin (- π/4) ].

w = acis(b)

a = √2

b = - π/4.

Answer 2

Q1)

b) The complex number z = 2 + 2 √3 i

The polar form of complex number z = x + iy is z = c [cos(d) + i sin(d)]

In this case x = 2, y = 2 √3

c = | z | = √ [2²  +(2√3)²]

c = √ (4 + 12)

c = √16

c = 4

Since x > 0, use the formula d = tan^-1(y/x)

d = tan^-1 (2 √3/2)

d = tan^-1 (√3)

d = 60

d =  π/3

Polar form of z = 2 + 2 √3 i is 4 [cos ( π/3) + i sin ( π/3) ].

Eulers formula e^iθ = cos(θ) + i sin(θ)

z = ce^di

We must express d in radians.

1° = π/180

So 60° = 60π/180

           = π/3

           = 1.047 radians

So we can write 4(cos60° + i sin 60°)

= 4 e^{i π/3}

= 4 e^{1.047 i}

c = 4 and d = 1.047.