Clearly, the complex number 0=0+0i in H as a=b=0 and \(\displaystyle{a}{b}={0}\ge{0}.\)

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 \(\displaystyle{a}{b}={0}\ge{0}\)

Hence, \(\displaystyle{1}+{0}{i}\in{H}\)

Consider the complex number 0−i.

For the complex number 0−i, a=0, b=−1 and \(\displaystyle{a}{b}={0}\ge{0}.\)

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

Hence, \(\displaystyle{1}−{i}\notin{H}.\)

Thus, \(\displaystyle{1}+{0}{i},{0}−{i}\in{H}\), but their \(\displaystyle\sum{1}−{i}!{n}{H}.\)

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 \(\displaystyle{a}{b}={0}\ge{0}\)

Hence, \(\displaystyle{1}+{0}{i}\in{H}\)

Consider the complex number 0−i.

For the complex number 0−i, a=0, b=−1 and \(\displaystyle{a}{b}={0}\ge{0}.\)

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

Hence, \(\displaystyle{1}−{i}\notin{H}.\)

Thus, \(\displaystyle{1}+{0}{i},{0}−{i}\in{H}\), but their \(\displaystyle\sum{1}−{i}!{n}{H}.\)

Therefore, H is not closed under addition.

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