Two function in one time

Hi everyone.This is my first post.I’m sorry for newby question but I have problem with invoke two function in one time.I was created client for RabbitMQ. While I was starting this script sequence of execute function depends on which function is implementation first in method START.Talking about features listen_for_messages(channel, queue_data.queue) and wait_for_message(user, channel).I want to both features “listen” in one time.Function wait_for_message it’s responsible for text which user write while listen_for_messages it’s responsible for recieve messages from another client. Please give me any suggestion because i have no idea how solve this problem.This is my CODE:

defmodule ElixirChat do

  def start do
    user = IO.gets("Type in your name: ") |> String.strip
    IO.puts "Hi #{user}, you just joined a chat room! Type your message in and press enter."

    {:ok, conn} = AMQP.Connection.open
    {:ok, channel} = AMQP.Channel.open(conn)
    {:ok, queue_data } = AMQP.Queue.declare(channel, "")

    AMQP.Exchange.fanout(channel, "super.chat")
    AMQP.Queue.bind(channel, queue_data.queue, "super.chat")

		**listen_for_messages(channel, queue_data.queue)**
**		wait_for_message(user, channel)**


		
    
  end
  
  def display_message(user, message) do
    IO.puts "#{user}: #{message}"
  end

  def wait_for_message(user, channel) do
     message = IO.gets("") |> String.strip
     publish_message(user, message, channel)
     wait_for_message(user, channel)
  end

  def listen_for_messages(channel, queue_name) do
	  AMQP.Basic.consume(channel,queue_name,nil,no_ack: true)
  	
  	receive do
  		{:basic_deliver, payload, _meta}->
  		{status, list} = JSON.decode(payload)
		  display_message(list["user"], list["message"])
		  listen_for_messages(channel, queue_name) 
    end	


		
  end

  def publish_message(user, message, channel) do
    { :ok, data } = JSON.encode([user: user, message: message])
    AMQP.Basic.publish channel, "super.chat", "", data
  end

end

ElixirChat.start