If z=r e^ i , then |e^ i z |=

Step-by-step solution

We have, aligned z &=r e^ i =r( +i ) i z &=i r( +i ) &=i r +i² r =-r +i r |e^ i z | &= |e^ (-r +i r ) | &= |e^ -r | |i r | &=e^ -r [ ²(r )+ ²(r ) ] &=e^ -r 1=e^ -r aligned