Question

represent in polar form z=cosa+sina+i (sina-cosa),a belongs (0,2pi)

Answer 1

For a complex no written as

z= x + iy

we know that it can be written as polar form as r(cos ∅ + isin∅)

where r = √x²+y²

and ∅ = tan⁻¹ y/x

Hence , when

z = (cosa + sina) + i (sina - cosa)

we have r = √(cosa+sina)² + (sina - cosa)²

= √(cos²a + sin²a + 2cosasina + sin²a + cos²a -2cosasina

=√1+1

=√2

To find ∅,

tan ∅ = (sina - cosa)/(cosa + sina)

=> sin ∅/ cos∅ =  (sina - cosa)/(cosa + sina)

cross multiplying and arranging we get ,

sin∅cosa - cos∅sina = -cos∅cosa + sin∅sina

=>sin(∅-a) = - cos(∅-a)

=> tan(∅-a) = -1

=> ∅-a = 3π/4 or 7π/4

=> ∅ = 3π/4 + a  or 7π/4 + a

Hence z can be written as ,

z= √2(cos(3π/4 + a) + i sin(3π/4 + a))

      or

z = √2(cos(7π/4 + a) + i sin(7π/4 + a))