Determine whether H is a subgroup of the complex numbers C with addition

$H=\{a+bi\mid a,b\in R,ab\ge 0\}$

Annette Arroyo
2020-11-08
Answered

Determine whether H is a subgroup of the complex numbers C with addition

$H=\{a+bi\mid a,b\in R,ab\ge 0\}$

You can still ask an expert for help

d2saint0

Answered 2020-11-09
Author has **89** answers

Clearly, the complex number 0=0+0i in H as a=b=0 and $ab=0\ge 0.$

So, H is a non empty subset of C.

In order for H to be a subgroup of C, H must be closed under addition.

Consider the complex number 1+0i.

For the complex number 1+0i, a=1, b=0 and$ab=0\ge 0$

Hence,$1+0i\in H$

Consider the complex number 0−i.

For the complex number 0−i, a=0, b=−1 and$ab=0\ge 0.$

Hence, 0−i in H.

Now, (1+0i)+(0−i)=(1+0)+(0−1)i=1−i.

For the complex number 1−i, a=1, b=−1 and ab=−1<0.

Hence,$1-i\notin H.$

Thus,$1+0i,0-i\in H$ , but their $\sum 1-i!nH.$

Therefore, H is not closed under addition.

So, H is not a subgroup of C under addition.

So, H is a non empty subset of C.

In order for H to be a subgroup of C, H must be closed under addition.

Consider the complex number 1+0i.

For the complex number 1+0i, a=1, b=0 and

Hence,

Consider the complex number 0−i.

For the complex number 0−i, a=0, b=−1 and

Hence, 0−i in H.

Now, (1+0i)+(0−i)=(1+0)+(0−1)i=1−i.

For the complex number 1−i, a=1, b=−1 and ab=−1<0.

Hence,

Thus,

Therefore, H is not closed under addition.

So, H is not a subgroup of C under addition.

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