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.

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  • 9 months ago

    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,

    Dragon.Jade :-)

  • MyRank
    Lv 6
    10 months ago

    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))

  • 10 months ago

    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

  • 10 months ago

    Know your unit circle

    cos(π/4) = √2 / 2

    sin(π/4) = √2 / 2

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  • 10 months ago

    You win the fried brain award for today.

    a + bi

    a = 2cos(pi/4)

    b = 2sin(pi/4)

    Take a step back. Drink water. Rest

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