# B Representation of complex of square root of negative i with unitary power.

1. Aug 25, 2017

### Leo Authersh

Can $sqrt(-i)$ be expressed as a complex number z = x + iy with unitary power?

2. Aug 25, 2017

### Staff: Mentor

What does "with unitary power" mean?
The complex square root has two solutions, both of them are complex numbers, they can be expressed as z=x+iy like every complex number.

3. Aug 25, 2017

### Leo Authersh

If any unit real number when rooted, powered, multiplied and divided gives a complex unit, then how can an unit imaginary number when rooted equates to a fractional complex number?

For example, $sqrt(i)$ equates to a non unit value i.e. 0.707+0.707i (approx)

4. Aug 25, 2017

### vanhees71

Well, your number is $\exp(\mathrm{i} \pi/4)=(1+\mathrm{i})/\sqrt{2}$ and thus its modulus is 1 as it must be.

5. Aug 25, 2017

### Leo Authersh

@vanhees71 Thank you, now I understood that 0.707+0.707i represents the x,y coordinate in the complex plan and not a value.

But shall we simply write as 0.707,0.707i so that it might not be misinterpreted as arithmetically additive?

Last edited by a moderator: Aug 25, 2017
6. Aug 25, 2017

### vanhees71

The point of complex numbers is that you can calculate with them as with real numbers, because they obey all the axioms of a field concerning the fundamental arithmethics of + and $\times$.

At the same time of course you can interpret real and imaginary part of a complex number as Cartesian coordinates in an Euclidean plane (Gauss's plane of numbers). This has great advantages, because it simplifies standard tasks like, e.g., rotations. The rotation of a vector $\vec{x}=(x,y)$ can be very easily calculated by writing $z=x+\mathrm{i} y$ and then the rotated vector is given by
$$z'=\exp(\mathrm{i} \phi) z,$$
where $\phi \in \mathbb{R}$ is the rotation angle (in radians) which you can easily check by using
$$\exp(\mathrm{i} \phi)=\cos \phi+\mathrm{i} \sin \phi$$
and multiplying out the product, splitting it again in real and imaginary part.

7. Aug 25, 2017

### Leo Authersh

After from my previous two questions, I have gained some more understanding of the complex numbers and now I can see this question arises out of my misinterpretation of complex number. Thank you.

8. Aug 25, 2017