Root of iota plus root of minus iota
I assume by iota you mean i, which is √-1
√i = (1+i)/√2
√-i = (1-i)/√2
add them to get 2/√2 = √2
Well, if we dig deep into the world of complex numbers, we can solve this equation using a bit of clownish math. Let's assume that "root" refers to square root.
Now, the square root of -1 is represented by the imaginary unit "i." But when we try to take the square root of i, things get a little tricky. It actually has multiple solutions!
So, the square root of i can be written as (+/-) (√2/2 + √2/2i). Let's call this solution "A."
Similarly, the square root of -i can also be written as (+/-) (√2/2 - √2/2i). Let's call this solution "B."
Now, if we sum solution A and B, we get (√2/2 + √2/2i) + (√2/2 - √2/2i). The imaginary parts (√2/2i and -√2/2i) will cancel each other out, leaving us with (√2/2 + √2/2). Simplifying further, we get √2.
So, the root of iota plus root of -iota is √2, which proves that even in the world of complex numbers, clowns can make math a little more amusing!
To find the value of √i + √(-i), we can start by expressing both square roots in terms of their polar form.
Let's start with √i:
i can be written as i = cos(π/2) + isin(π/2) in polar form.
Taking the square root of i, we have:
√i = (√r)(cos(θ/2) + isin(θ/2))
For i, r = 1, and θ = π/2, so:
√i = (√1)(cos(π/2)/2 + isin(π/2)/2)
= (1)(cos(π/4) + isin(π/4))
= cos(π/4) + isin(π/4)
= √2/2 + √2/2i
Next, let's move on to √(-i):
(-i) can be written as (-i) = cos(3π/2) + isin(3π/2) in polar form.
Taking the square root of -i, we have:
√(-i) = (√r)(cos(θ/2) + isin(θ/2))
For -i, r = 1, and θ = 3π/2, so:
√(-i) = (√1)(cos(3π/4)/2 + isin(3π/4)/2)
= (1)(cos(3π/8) + isin(3π/8))
= cos(3π/8) + isin(3π/8)
Now, to find the sum of these two square roots:
√i + √(-i) = (√2/2 + √2/2i) + (cos(3π/8) + isin(3π/8))
Adding real parts and imaginary parts separately:
= (√2/2 + cos(3π/8)) + (√2/2i + isin(3π/8))
Simplifying further:
= √2/2 + cos(3π/8) + (isin(3π/8) + √2/2i)
Combining the real and imaginary parts:
= √2/2 + cos(3π/8) + √2/2i + isin(3π/8)
The final answer is √2/2 + cos(3π/8) + √2/2i + isin(3π/8).
To find the value of the expression √i + √(-i), we need to determine the square roots of i and -i.
Since i is an imaginary unit, we can write it as i = √(-1).
Now, let's calculate the square roots of i and -i step by step:
1. Square root of i:
To find the square root of i, we can rewrite i using its polar form. The polar form of i is i = e^(iπ/2), where e is Euler's number. The square root of i can be found by taking the square root of its magnitude (1) and halving its argument (π/4).
Using the polar form, the square root of i is:
√i = √(e^(iπ/2)) = e^(iπ/4)
2. Square root of -i:
Similarly, we can write -i = √(-1) = e^(iπ). To find the square root of -i, we again take the square root of its magnitude (1) and halve its argument (π/2).
Using the polar form, the square root of -i is:
√(-i) = √(e^(iπ)) = e^(iπ/2)
Now, substituting the values of the square roots back into the original expression:
√i + √(-i) = e^(iπ/4) + e^(iπ/2)
At this point, the expression cannot be simplified further, as it involves complex exponential functions. If you need the numerical value of √i + √(-i), you can use a scientific calculator or a math software.
Let root iota &root -iota = y
Now square on both side we get i - i+ 2 i^1/2 -i^1/2
= 2 [-i^2]^1/2
= 2
As y^2 = 2
Then y 2^1/2