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- #1

#### lemonthree

##### New member

- Apr 11, 2016

- 20

"The map

Is this true or false? I'm guessing it's true because

φ (j) = | j |, which means

φ (j * k) = | j * k |

=| j | * | k |

= φ ( j ) * φ ( k ).

- Thread starter lemonthree
- Start date

- Thread starter
- #1

- Apr 11, 2016

- 20

"The map

Is this true or false? I'm guessing it's true because

φ (j) = | j |, which means

φ (j * k) = | j * k |

=| j | * | k |

= φ ( j ) * φ ( k ).

- Oct 18, 2017

- 248

That is true, but you should also check that

- $\phi(x^{-1}) = \phi(x)^{-1}$
- $\phi(1) = 1$