Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Aug 25, 2017 #1
    Can ##sqrt(-i)## be expressed as a complex number z = x + iy with unitary power?
     
  2. jcsd
  3. Aug 25, 2017 #2

    mfb

    User Avatar
    2017 Award

    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.
     
  4. Aug 25, 2017 #3
    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)
     
  5. Aug 25, 2017 #4

    vanhees71

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Well, your number is ##\exp(\mathrm{i} \pi/4)=(1+\mathrm{i})/\sqrt{2}## and thus its modulus is 1 as it must be.
     
  6. Aug 25, 2017 #5
    @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
  7. Aug 25, 2017 #6

    vanhees71

    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    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.
     
  8. Aug 25, 2017 #7
    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.
     
  9. Aug 25, 2017 #8

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted