2021/112) Let G= $\{a+bi$ in complex: $a^2 +b^2=1\}$. Is G a group under multiplication?

For it to be group folllowing must be tue.

1) it should be closed

that is if  x = a + ib and y = c + id and $(a^2+b^2) = 1$ and $c^2+d^2=1$ 

and xy = m + ni then $m^2+n^2 =1$

we have $xy = m + ni = (a+ib)(c+id) = (ac - bd) + (bc + ad)i$

we have m = ac - bd and n = bc + ad

$m^2 + n^2 = (ac-bd)^2 + (bc + ad)^2 = a^2c^2 - 2abcbd + b^2d^2 + b^2 c^2 + 2abcd + a^2d^2$

 $= a^2c^2 + b^2d^2 + b^2c^2 + a^2d^2 = (a^2+b^2)(c^2 + d^2) = 1$ 

So it is closed 

2) It should have an identity

1 or 1+0i is identity element as $(a+bi)(1+0i) = a+ bi$

3) it should have an inverse  

because $a^2+b^2=1$ so it it not zero and hence it has inverse and we need to show that if

m+in is inverse then $m^2+ n^2 =1$ that is the inverse is in this group

$m + in = \frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a=ib)} = \frac{a-ib}{a^2+b^2} = a - ib$

'so m = a , n = - b and $m^2 + n^2 = a^2 + (-b)^2 = a^2 + b^2 =1$

so it has an inverse

4) assosiativity law holds as unde rcomplex number multiplication assosiativity holds