# Convert Complex Numbers to Rectangular Form.?

How do I continue with 8a? I stopped at 2(cos(pi/4)+isin(pi/4)). But then, I dont understand how to continue afterward. Relevance
• Hello,

The required knowledge:

𝑒ⁱᶿ = cos(𝜃) + 𝑖·sin(𝜃)

cos(-𝜃) = cos(𝜃)

sin(-𝜃) = -sin(𝜃)

cos(𝜋/4) = sin(𝜋/4) = (√2)/2

cos(𝜋/3) = ½

sin(𝜋/3) = (√3)/2

From this, you get:

2·𝑒^(𝑖𝜋/4)

= 2·[cos(𝜋/4) + 𝑖·sin(𝜋/4)]

= 2·[(√2)/2 + 𝑖·(√2)/2]

= (√2) + 𝑖·(√2)    ◄◄◄(a)

5·𝑒^(𝑖𝜋/3)

= 5·[cos(𝜋/3) + 𝑖·sin(𝜋/3)]

= 5·[½ + 𝑖·(√3)/2]

= (5/2) + 5𝑖·(√3)/2    ◄◄◄(b)

[𝑒^(-𝑖𝜋/3)] / 3

= [cos(-𝜋/3) + 𝑖·sin(-𝜋/3)] / 3

= [cos(𝜋/3) − 𝑖·sin(𝜋/3)] / 3

= [½ − 𝑖·(√3)/2] / 3

= (1/6) − 𝑖·(√3)/6    ◄◄◄(c)

𝜋·𝑒^(-𝑖𝜋/4)   = 𝜋·[cos(-𝜋/4) + 𝑖·sin(-𝜋/4)]

= 𝜋·[cos(𝜋/4) − 𝑖·sin(𝜋/4)]

= 𝜋·[(√2)/2 − 𝑖·(√2)/2]

= 𝜋(√2)/2 − 𝑖·𝜋(√2)/2   ◄◄◄(d)

Regards,

• Log in to reply to the answers
• 2(cos(π/4)+isin(π/4))

e^(iθ) = cos θ + i sin θ

where θ = π/4, r = 2

2(cos(π/4)+isin(π/4)) = 2(e^(π/4i))

• Log in to reply to the answers
• r e^(iθ) = r cis(θ) = r cosθ + i r sinθ

cos(π/4) = sin(π/4) = √½, so (a) 2 e^(i π/4) = 2√½ + 2√½ i = √2 + √2 i

• Log in to reply to the answers

cos(π/4) = √2 / 2

sin(π/4) = √2 / 2

• Log in to reply to the answers
• Log in to reply to the answers