Results 1 to 9 of 9

Math Help - Determine if this set of complex numbers forms a subgroup

  1. #1
    Member
    Joined
    Dec 2010
    Posts
    92

    Determine if this set of complex numbers forms a subgroup

    I think I answered this correctly, but I wanted to be sure.

    Problem:
    Determine whether the given subset H of the group G is a subgroup.

    G = \mathbb{C}^{x}, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

    H = \left\{z \in \mathbb{C} \hspace{2mm} |\hspace{2mm} |z| = 1\right\}

    ******************

    For H to be a subgroup of G, the following three things must be satisfied:
    1) Closed under the binary operation of G (in this case, multiplication of complex numbers)
    2) H has the identity element of G
    3) For all a \in H, there exists an a^{-1} \in H.

    |z| is defined as |z|=\sqrt{a^2+b^2} where a and b come from the general complex number form a+bi.

    Since |z| = 1, it must be true that only one of the pair a and b is 1, and the other needs to be zero.

    Therefore, this set H only has four elements:

    1+0i
    0+1i
    -1+0i
    0-1i

    1) Multiplying two elements of this set will always return a result that is still within the set H, and therefore the set H is closed under the binary operation. Since there are always going to be two zero terms when foiling two of these terms, only the terms where +/-1 are multiplied together, always returning one of the four elements in the group.

    2) An identity of group G will satisfy

    a*e = e*a = a

    In this case, e=1+0i intuitively satisfies this, and is the unique identity element of group G. |e|=|1+0i|=1, so the set H has the identity of G.

    3) The inverse of a complex number is the conjugate. So a complex number x = a+bi has the inverse x^{-1}=a-bi. Looking at the list of four elements from part (1), it can be seen that the first two elements are x and y, and the following two are the conjugates (inverses) of x and y, and vice versa. Therefore, every element in H has an inverse in H.

    Since (1), (2), and (3) are satisfied, H is a subgroup of G.

    Am I correct? Is there anything here that I did wrong?
    Last edited by tangibleLime; October 11th 2011 at 11:29 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member ModusPonens's Avatar
    Joined
    Aug 2010
    Posts
    125
    Thanks
    14

    Re: Determine if this set of complex numbers forms a subgroup

    No, it's wrong. The elements are the circunference of points at the distance 1 from the point (0,0). They are of the form e^ix. try it again. Apart from that mistake, you had everything right.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,383
    Thanks
    750

    Re: Determine if this set of complex numbers forms a subgroup

    Quote Originally Posted by tangibleLime View Post
    I think I answered this correctly, but I wanted to be sure.

    Problem: Determine whether the given subset H of the group G is a subgroup.

    G = \mathbb{C}^{x}, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

    H = \left\{z \in \mathbb{C} \hspace{2mm} |\hspace{2mm} |z| = 1\right\}

    ******************

    For H to be a subgroup of G, the following three things must be satisfied:
    1) Closed under the binary operation of G (in this case, multiplication of complex numbers)
    2) H has the identity element of G
    3) For all a \in H, there exists an a^{-1} \in H.
    just a remark, here, that 1) & 3) together imply 2), so you don't need to show this. (if ab is in H when a,b are, and a^-1 is in H, then taking b = a^-1 proves 2)).

    |z| is defined as |z|=\sqrt{a^2+b^2} where a and b come from the general complex number form a+bi.

    Since |z| = 1, it must be true that only one of the pair a and b is 1, and the other needs to be zero.
    this is false. √2/2 + i√2/2 is also in H, as is easily verified.

    Therefore, this set H only has four elements:

    1+0i
    0+1i
    -1+0i
    0-1i

    1) Multiplying two elements of this set will always return a result that is still within the set H, and therefore the set H is closed under the binary operation. Since there are always going to be two zero terms when foiling two of these terms, only the terms where +/-1 are multiplied together, always returning one of the four elements in the group.
    all useless.

    2) An identity of group G will satisfy

    a*e = e*a = a

    In this case, e=1+0i intuitively satisfies this, and is the unique identity element of group G. |e|=|1+0i|=1, so the set H has the identity of G.

    3) The inverse of a complex number is the conjugate. So a complex number x = a+bi has the inverse x^{-1}=a-bi. Looking at the list of four elements from part (1), it can be seen that the first two elements are x and y, and the following two are the conjugates (inverses) of x and y, and vice versa. Therefore, every element in H has an inverse in H.

    Since (1), (2), and (3) are satisfied, H is a subgroup of G.

    Am I correct? Is there anything here that I did wrong?
    the proper approach here, is to first show that |zw| = |z||w|, for any 2 complex numbers z = a+ib, w = c+id.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2010
    Posts
    92

    Re: Determine if this set of complex numbers forms a subgroup

    Quote Originally Posted by ModusPonens View Post
    No, it's wrong. The elements are the circunference of points at the distance 1 from the point (0,0). They are of the form e^ix. try it again. Apart from that mistake, you had everything right.
    I don't quite understand this. I thought that |z|, where z is a complex number, was the modulus of z and defined as sqrt(x^2+y^2). Why is it instead e^ix? That forms the roots of unity, correct?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,383
    Thanks
    750

    Re: Determine if this set of complex numbers forms a subgroup

    suppose z = a+ib.

    then |z| = \sqrt{a^2+b^2} so

    |z|^2 = a^2+b^2

    if |z| = 1, we obtain the equation:

    a^2+b^2 = 1. for which pairs (a,b) is this true? this equation should ring a bell....
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2010
    Posts
    92

    Re: Determine if this set of complex numbers forms a subgroup

    Quote Originally Posted by Deveno View Post
    suppose z = a+ib.

    then |z| = \sqrt{a^2+b^2} so

    |z|^2 = a^2+b^2

    if |z| = 1, we obtain the equation:

    a^2+b^2 = 1. for which pairs (a,b) is this true? this equation should ring a bell....
    Isn't that true for (1,0), (-1,0), (0,1), (0,-1) like I said? Why am I supposed to bring in e^ix? (from what ModusPonens said)

    Edit: I see that those are just integer solutions... but I still don't see where e^ix is coming from.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1

    Re: Determine if this set of complex numbers forms a subgroup

    Quote Originally Posted by tangibleLime View Post
    Problem: [/B]Determine whether the given subset H of the group G is a subgroup.
    G = \mathbb{C}^{x}, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.

    H = \left\{z \in \mathbb{C} :\hspace{2mm} |z| = 1\right\}
    The set H is the whole unit circle in the complex plane.
    Here is a neat theorem: If H is a non-empty subset of a group G then H is subgroup if and only if for \{x,y\}\subset H it is true that xy^{-1}\in H.

    For this H if y\in H then y^{-1}=\overline{y}, the conjugate, and |\overline{y}|=1 .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2010
    Posts
    92

    Re: Determine if this set of complex numbers forms a subgroup

    As soon as you posted, I realized how I was thinking about it the wrong way. H is the set of complex numbers on the unit circle SUCH THAT |z|=1. Okay. Thanks
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,383
    Thanks
    750

    Re: Determine if this set of complex numbers forms a subgroup

    e^{iy} = cos(y) + i sin(y) by definition.

    so what is |e^{iy}| ?

    by direct calculation:

    |e^{iy}| = \sqrt{cos^2(y) + sin^2(y)} = \sqrt{1} = 1

    now, let's show that every complex number z with |z| = 1, can be written in this form:

    let z = a+ib, and define t = arctan(b/a),

    note that cos t = cos(arctan(b/a)) = \frac{a}{\sqrt{a^2+b^2}}

    and that sin t = sin(arctan(b/a)) = \frac{b}{\sqrt{a^2+b^2}}

    since \sqrt{a^2+b^2} = 1,

    cos t = a, sin t = b, and z = a+bi = cos t + i sin t.

    one caveat: a might be 0, in which case arctan(b/a) is undefined.

    in that case, use t = \pi/2, if z is in the top half-plane, or

    t = -\pi/2, if z is in the lower half-plane.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Polar Forms, Sin and Cos Angles
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: September 21st 2011, 02:11 PM
  2. Replies: 2
    Last Post: May 16th 2011, 01:09 PM
  3. complex numbers - product of polar forms
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: April 10th 2011, 01:13 PM
  4. Complex Exponential Forms
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: February 14th 2010, 03:34 AM
  5. Replies: 1
    Last Post: May 24th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum