# 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.

### 5 Answers

- Dragon.JadeLv 79 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 :-)

- MyRankLv 610 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))

Source(s): http://myrank.co.in/ - Φ² = Φ+1Lv 710 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

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- RealProLv 710 months ago
You win the fried brain award for today.

a + bi

a = 2cos(pi/4)

b = 2sin(pi/4)

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