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Show that the set M of complex numbers z with the condition | z | = 1 forms a group with respect to the operation of multiplication of complex numbers.

1 Answer

  • A non-empty set G, together with an operation $*$ i.e., $(G, *)$ is said to be a group if it satisfies the following axioms
  • (1) Closure axiom : $a,b\in G\Rightarrow a*b\in G$
  • (2) Associative axiom : $\forall a, b, c \in G, (a * b) * c = a * (b * c)$
  • (3) Identity axiom : There exists an element $e \in G$ such that $a * e = e * a = a, \forall a ^{-1} G.$
  • (4) Inverse axiom : $\forall a \in G$ there exists an element $a^{-1}\in G$ such that $a^{-1}*a=a*a^{-1}=e.$
  • e is called the identity element of $G$ and $a^{-1}$ is called the inverse of $'a'$ in $G.$
Step 1:
Let $M=\{z\in c\mid z\mid =1\}$ where $c$ is set of complex numbers.
Closure :
The product of $z_1,z_2\in M$ is $z_1,z_2\in M$ since $\mid z_1z_2\mid=\mid z_1\mid\mid z_2\mid=1$
Closure is satisfied.
Step 2:
Associativity :
Multiplication of complex numbers is associative.
Step 3:
Existence of identity :
$1=1+0i\in M$
Once $\mid 1\mid=1$
$\therefore $the identity exists.
Step 4:
Existence of inverse :
For $z\in M,z^{-1}=\large\frac{\bar{z}}{\mid z\mid}$$=\bar{z}$
$\mid \bar{z}\mid =\mid z\mid=1$
$\therefore z^{-1}\in M$
Therefore every element in $M$ has an inverse.
The four group axioms being satisfied.
$M$ is a group under multiplication.
answered Sep 14, 2013 by sreemathi.v

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