# lecture8 - 5 The Exponential Function and its Friends The...

This preview shows pages 1–2. Sign up to view the full content.

5. The Exponential Function and its Friends The exponential function. Define exp z = n =0 z n n ! . This has radius of convergence R where 1 /R = lim sup n →∞ (1 /n !) 1 /n = 0 , so R = and the series converges absolutely for all z C . We may differentiate term by term (Theorem!) to get d dz exp z = n =1 n z n 1 n ! = n =1 z n 1 ( n 1)! = n =0 z n n ! = exp z. Exponential of a sum. Set f ( z ) = exp z exp( c z ), c C . Then f ( z ) = d dz exp z exp( c z )+exp z d dz exp( c z ) = exp z exp( c z ) exp z exp( c z ) = 0 . By Theorem 3.5, this tells us f ( z ) is constant. This constant must be f (0) = exp c , so exp z exp( c z ) = exp c . Set c = z 1 + z 2 , z = z 1 , then exp( z 1 + z 2 ) = exp z 1 exp z 2 . From this it is easy to deduce (induction) that for n Z + , exp nz = (exp z ) n . The number e . Define e = exp 1(= 2 . 7182818 . . . ). The above shows exp n = e n , n Z + . Clearly exp 0 = 1 = e 0 . Hence exp n exp( n ) = exp( n n ) = exp 0 = 1 so exp( n ) = (exp n ) 1 = e n . So exp( n ) = e n , n Z . For a rational number m/n ( n > 0), (exp( m/n )) n = exp( nm/n ) = exp m = e m , so exp( m/n ) = ( e m ) 1 /n = e m/n . It is then consistent with this to use the notation exp z = e z , for all z C . The formula (*) becomes e z 1 + z 2 = e z 1 e z 2 . If z = x + iy , we have e z = e x e iy . Notice also that exp z exp( z ) = exp 0 = 1 , so exp

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern