What is the Value of √i + √-i?

kini.Amith
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Homework Statement


The value of √i + √-i , where i=√-1 is
(a) 0 (b) 1/√2 (c) √2 (d) -√2

Only one option can be chosen

Homework Equations


The Attempt at a Solution



Let (x + iy)2=i
Solving for x and y, I got
√i = +1/√2 (1+i) or -1/√2 (1+i)

Similarly i got
√-i = +1/√2 (1-i) or -1/√2 (1-i)

Now how do I calculate, √i - √-1,
i.e since there 2 possible values for each √i and √-i , which do i subtract from which?

Using another method I got the final nswer as + or - √2 , but u can only choose one of the given options
 
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They are asking for the principal square root.
 
D H said:
They are asking for the principal square root.
How are principal square roots defined for imaginary numbers?
For eg, if you have 1 - i and -1 + i as the 2 square roots, which do you choose as the principal one?
If you have i and -i, which do you choose?
 
Have you tried looking up the definition in your textbook or notes?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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