d4nte asked in Science & MathematicsMathematics · 1 month ago

# 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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• 1 month 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,

• MyRank
Lv 6
1 month 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))

• 1 month 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

• 1 month ago

cos(π/4) = √2 / 2

sin(π/4) = √2 / 2

• 1 month ago

You win the fried brain award for today.

a + bi

a = 2cos(pi/4)

b = 2sin(pi/4)

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