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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

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    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

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    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: Aug 25, 2017
  7. Aug 25, 2017 #6

    vanhees71

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    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

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