8/12/2019 Equacoes Resolvidas de Mat

1/2

8/12/2019 Equacoes Resolvidas de Mat

2/2

7. As log(1) = i and log 11

= log(1) = i=i = log(1) this result is not true

for all z C.

8. The equation cosh(z) =2 is satisfied if and only if 12(ez +ez) =2, that is if and onlyifez +ez =4, or if and only ife2z + 4ez + 1 = 0. Therefore

ez =4 12

2 =2

3.

Let 1 =2 3, 2 =2 + 3, so we now have to solve the equations ez = 1 andez = 2. Note that both 1 and 2 are negative so that the polar form of both is |j|ei.Solving the equation ez = j as question 1 we get the set of solutions of cosh(z) =2 to be{log(|j|) +i(2n+ 1): j = 1, 2, n Z}.

9. Using the estimation propertyC(0,1)

sin(z)

z2 dz

C(0, 1) sup{sin(z)z2

:zC(0, 1)}= 2sup{

sin(z)

z2

:|z|= 1}

= 2

sup{| sin(z

)|:|z|= 1} 2cosh(1)

2eas if|z|= 1 then|mz| 1 so| sin(z)| cosh(mz)cosh(1).