what is (i+1)^(1/2)?

Answer 1

Use DeMoivre's Theorem:

z^(1/n) = (r(cos(x) + i sin(x)))^(1/n)

= r^(1/n)[cos((x + 2kπ)/n) + i sin((x + 2kπ)/n)], 0 < k < n - 1

Which is ugly-looking but it's not too bad.

First, write (i + 1)^(1/2) in polar coordinates:

r² = 1² + 1² = 2, so r = √2
θ = arctan(1/1) = π/4

So (1 + i) is √2(cos(π/4) + i sin(π/4))

Then:

(1 + i)^(1/2) = (√2)^(1/2)(cos((π/4 + 2kπ)/2) + i sin((π/4 + 2kπ)/2))

= 2^(1/4)(cos(π/8 + kπ) + i sin(π/8 + kπ))

So for k = 0, you'd have the first root:

2^(1/4)(cos(π/8) + i sin(π/8)) ≈ 1.1 + 0.455i

And for k = 1, you'd have the second root:

2^(1/4)(cos(π/8 + π) + i sin(π/8 + π)) ≈ -1.1 - 0.455i

Answer 2

It is the sqrt of(1+i)
To find roots (any) of complex numbers it is easier to work in polar form

1+i = r The sqare root would be sqrtr
With further values of k you would return to the same points

The result is 2^1/4 <22.5 degrees and 2^1/4 <202.5 degree.
Then if you wish or need you can return to binomial writing of complex numbers

Answer 3

The angle θ associated with 1 + i is π/4. So the angle associated with the square root is half that or π/8. Using the half angle formula for sin(π/8) we have:

sin(π/8) = √{[1 - cos(π/4)]/2} = √{[1 - 1/√2]/2} = √{[2 - √2]/4}
sin(π/8) = √(2 - √2)/2

cos(π/8) = √{[1 + cos(π/4)]/2} = √{[1 + 1/√2]/2} = √{[2 + √2]/4}
cos(π/8) = √(2 + √2)/2

(1 + i) = (√2){cos(π/4) + i sin(π/4)}
(1 + i)^(1/2) = ±(2)^(1/4){cos(π/8) + i sin(π/8)}
(1 + i)^(1/2) = ±(2)^(1/4){√(2 + √2)/2 + i√(2 - √2)/2}

Answer 4

Z=1+Ä°
r=sqrt(2)
angle=45
z=sqrt(2)(cos45+2 k pi)+isin(45+2 k pi)

for k=0
root1=2(cos22,5+isin22,5)

for k=1
root2=2(cos(22,5+180)+isin(22,5+180)