Sorry, but you're wrong. We know i = e^(πi/2) so √i = e^(πi/4) = cos(π/4) + i*sin(π/4) = (1 + i)/√2. Similarly -i = e^(-πi/2), so √(-i) = (1 - i)/√2. Hence √i + √(-i) = 2/√2 = √2. There is no ± because √ denotes the principal value of the square root which is the value whose complex argument lies in the semi-open interval (-π/2, π/2].
There should not be plus minus sign, the plus minus arises when you take a square root, for a signed square root it is always the sign of square root, in this case only positive value.
No quite "the positive value" when dealing with complex numbers. The principal square root of a complex number is generally taken to be the value whose complex argument lies in the semi-open interval (-π/2, π/2], which makes the complex square root function compatible with the real square root function.
Insufficient information. When taking the square root of a complex number, z = r e^(iθ), it will be √z = √r e^(iθ/2), and you have to decide on a "cut-line" for the argument (θ). There are two common choices: [0, 2π), and (-π, π]. This will not matter for √i = (1 + i)/√2; but will make a difference for √-i. Under the first choice, -i has argument, θ = 3π/2, and √-i = (-1 + i)/√2; so √i + √-i = i√2. Under the second choice, -i has argument, θ = -π/2, and √-i = (1 - i)/√2; so √i + √-i = √2. Fred
Hey bro which whiteboard app you use plz tell ❤❤❤❤
Sqrt[-i]+Sqrt[i]=Sqrt[2] It’s in my head.
But you could have just let the equation equal to x and square both sides...will get same ans
You can't let the equation to be real part(x) alone but you could use x+yi
@catch-uptutors2691 ooh don't get confused by taking x as a real part...let x just a complex no ...at last it's just a variable
You beat me to it. I agree with you.
You all should check the comment action and see truly this man is a mathematician
√i + √-i = x
√i + √-1√i = x
√i + i√i = x
√i(1 + i) = x
i(1 + 2i - 1) = x^2
2i^2 = x^2
-2 = x^2
√-2 = x
2i = x
Why is this wrong?
why not start with x = √i + √-1, square both sides and solve for x? That's a lot faster.
±i=e^{±iπ/2}; √(±i)=√(e^{±iπ/2})=e^{±iπ/4}=cos(±π/4)+i*sin(±π/4)=cos(π/4)±i*sin(π/4)=√2/2±i√2/2.
Thus, √i+√(-i)=√2/2+i√2/2+√2/2-i√2/2=2(√2/2)=√2
±V2, ±V2i.
Input
sqrt(-i) + sqrt(i) = sqrt(2)
Exact result
(-1)^(1/4) - (-1)^(3/4) = sqrt(2)
True
I got 4 solutions... let p^2=1=q^2. sqrt(i)+sqrt(-i) = sqrt(i)*(1+p*i) = q*(1+i)/sqrt(2)*(1+p*i) = q*(1-p+i*(1+p))/sqrt(2)
Sorry, but you're wrong.
We know i = e^(πi/2) so √i = e^(πi/4) = cos(π/4) + i*sin(π/4) = (1 + i)/√2. Similarly -i = e^(-πi/2), so √(-i) = (1 - i)/√2. Hence √i + √(-i) = 2/√2 = √2.
There is no ± because √ denotes the principal value of the square root which is the value whose complex argument lies in the semi-open interval (-π/2, π/2].
There should not be plus minus sign, the plus minus arises when you take a square root, for a signed square root it is always the sign of square root, in this case only positive value.
No quite "the positive value" when dealing with complex numbers.
The principal square root of a complex number is generally taken to be the value whose complex argument lies in the semi-open interval (-π/2, π/2], which makes the complex square root function compatible with the real square root function.
at 3.5 you made a mistake +1^2 does not equal -1
i^2=-1 not 1^2
Insufficient information.
When taking the square root of a complex number, z = r e^(iθ), it will be √z = √r e^(iθ/2), and you have to decide on a "cut-line" for the argument (θ).
There are two common choices: [0, 2π), and (-π, π]. This will not matter for √i = (1 + i)/√2; but will make a difference for √-i.
Under the first choice, -i has argument, θ = 3π/2, and √-i = (-1 + i)/√2; so √i + √-i = i√2.
Under the second choice, -i has argument, θ = -π/2, and √-i = (1 - i)/√2; so √i + √-i = √2.
Fred