Question

Convert,the complex number into polar form: 4+4i

Answer 1

z = a + bi
z = 4 + 4i

r² = a² + b²
r² = (4)² + (4)²
r² = 16 + 16
r² = 32
r = 4√(2)
r = 4(1.414)
r = 5.656


`cos\theta = (a)/(r)`

`cos\theta = (4)/(4√(2))`

`cos\theta = (4)/(4√(2)) * (√(2))/(√(2))`

`cos\theta = (4√(2))/(4√(4))`

`cos\theta = (4√(2))/(4(2))`

`cos\theta = (4√(2))/(8)`

`cos\theta = (√(2))/(2)`

`2(cos\theta) = 2((√(2))/(2))`

`2cos\theta = √(2)`

`2cos\theta = 1.414`


`sin\theta = (b)/(r)`

`sin\theta = (4)/(4√(2))`

`sin\theta = (4)/(4√(2)) * (√(2))/(√(2))`

`sin\theta = (4√(2))/(4√(4))`

`sin\theta = (4√(2))/(4(2))`

`sin\theta = (4√(2))/(8)`

`sin\theta = (√(2))/(2)`

`2(sin\theta) = 2((√(2))/(2))`

`2sin\theta = √(2)`

`2sin\theta = 1.414`

z = a + bi
z = rcosθ + (rsinθ)i
z = r(cosθ + i sinθ)

z = 4 + 4i
z = 5.656cosθ + (5.656sinθ)i
z = 5.656(cosθ + i sinθ)
z = 5.656(cos45 + i sin45)


`\theta = tan^(-1)(b)/(a)`

`\theta = tan^(-1)(4)/(4)`

`\theta = tan^(-1)(1)`

`\theta = 45`

The polar form of 4 + 4i is approximately equal to 5.656(cos45 + i sin45).