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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)