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write the Complex number in Polar Coordinates?
1) Evaluate (2+3i)^1/2
2) e^(2+pi/2i)
I know you use De moivres Theorem but when I get tan inverse of 3/2 it gives me a decimal point. So I obviously have to do something different but what?
Any help would be great thanks.
2 Answers
- RaffaeleLv 77 years ago
1)
√(2 + 3i)
go polar
x + iy = ρ(cos θ + i sin θ)
where
ρ = √(x^2 + y^2)
θ = arctan(y/x)
√(2 + 3i) = x + iy
square both sides
2 + 3i = x^2 + i^2 y^2 + 2ixy
2 + 3i = (x^2 - y^2) + (2xy) i
must be
x^2 - y^2 = 2
2xy = 3
from the second equation
y = 3/(2x)
substitute in the first
x^2 - 9/(4x^2) = 2
set x^2 = w
4w - 9/w = 8
4w^2 - 8w - 9 = 0
w = 1 - √13/2 ∨ w = √13/2 + 1
drop the negative root
x = √(√13/2 + 1)
y = 3/(2√(√13/2 + 1)) = √(√13/2 - 1)
√(2 + 3i) = √(√13/2 + 1) + i √(√13/2 - 1)
2)
e^(2+π i/2) = e^2 e^π i/2 = e^2 i
- 7 years ago
(2 + 3i)^(1/2)
a = 2
b = 3
sqrt(a^2 + b^2) = sqrt(2^2 + 3^2) = sqrt(13)
(sqrt(13) * (2/sqrt(13) + i * 3/sqrt(13)))^(1/2) =>
13^(1/4) * (cos(t) + i * sin(t))^(1/2)
cos(t) = 2/sqrt(13)
t = 0.98279372324732906798571061101467 + 2pi * k radians
13^(1/4) * (cos(0.983 + 2pi * k) + i * sin(0.983 + 2pi * k))^(1/2)
13^(1/4) * (cos(0.4914 + pi * k) + i * sin(0.4914 + pi * k))
13^(1/4) * (cos(0.4914) + i * sin(0.4914)) and 13^(1/4) * (cos(3.633) + i * sin(3.633))
(13^(1/4) , 0.4914)
(13^(1/4) , 3.633)
e^(2 + (pi/2) * i) =>
e^(2) * (cos(pi/2) + i * sin(pi/2)) =>
e^(2) * (0 + i * 1) =>
e^(2) * i
(e^(2) , pi/2)