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 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.
, the group of all nonzero complex numbers with operation given by multiplication of complex numbers.
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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 , there exists an .
is defined as where a and b come from the general complex number form .
Since , 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) 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 will satisfy
In this case, intuitively satisfies this, and is the unique identity element of group G. , so the set H has the identity of G.
3) The inverse of a complex number is the conjugate. So a complex number has the inverse . Looking at the list of four elements from part (1), it can be seen that the first two elements are and , and the following two are the conjugates (inverses) of and , 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?
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.
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)).
this is false. √2/2 + i√2/2 is also in H, as is easily verified.is defined as where a and b come from the general complex number form .
Since , it must be true that only one of the pair a and b is 1, and the other needs to be zero.
all useless.Therefore, this set H only has four elements:
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.
the proper approach here, is to first show that |zw| = |z||w|, for any 2 complex numbers z = a+ib, w = c+id.2) An identity of group will satisfy
In this case, intuitively satisfies this, and is the unique identity element of group G. , so the set H has the identity of G.
3) The inverse of a complex number is the conjugate. So a complex number has the inverse . Looking at the list of four elements from part (1), it can be seen that the first two elements are and , and the following two are the conjugates (inverses) of and , 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?
by definition.
so what is ?
by direct calculation:
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)) =
and that sin t = sin(arctan(b/a)) =
since ,
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 = , if z is in the top half-plane, or
t = , if z is in the lower half-plane.