SOLUTION: solve log(1+i) in the form of a+bi

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Question 1073714: solve log(1+i) in the form of a+bi

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

I assume that by log( ) you mean loge( ) and not 
log10( ) as most people in the US mean by log( ).  
When we want to indicate loge we write ln( ). If
you meant log10 then tell me in the thank-you note 
form below and I'll get back to you. 

Use Euler's equation:

         e%5E%28a%2Bbi%29%22%22=%22%22e%5Ea%28cos%28b%29%2Bi%2Asin%28b%29%29,

Substitute ln(1+i) for a+bi on the left 

         e%5E%28ln%281%2Bi%29%29%22%22=%22%22e%5Ea%28cos%28b%29%2Bi%2Asin%28b%29%29

         1%2Bi%22%22=%22%22e%5Ea%28cos%28b%29%2Bi%2Asin%28b%29%29

eq. 1    1%2Bi%22%22=%22%22e%5Ea%2Acos%28b%29%2Be%5Ea%2Ai%2Asin%28b%29%29

Set the real part on the left side of eq. 1 
equal to the real part on the right of eq. 1:

eq. 2    1%22%22=%22%22e%5Ea%2Acos%28b%29

Set the imaginary part on the left side of eq. 1 
equal to the imaginary part on the right of eq. 1:

eq. 3    1%22%22=%22%22e%5Ea%2Asin%28b%29

Divide equals by equals, that is, divide both side 
of eq. 2 by both sides of eq. 1

1%2F1%22%22=%22%22%28e%5Ea%2Asin%28b%29%29%2F%28e%5Ea%2Acos%28b%29%29%0D%0A%0D%0A%7B%7B%7B1%22%22=%22%22sin%28b%29%2Fcos%28b%29

1%22%22=%22%22tan%28b%29

The tangent is positive in QI and QIII so

b%22%22=%22%22matrix%281%2C3%2Cpi%2F4%2Cor%2C5pi%2F4%29

However in eq. 3

1%22%22=%22%22e%5Ea%2Asin%28b%29,

the left side is positive, and ea is positive,
so we can discard the QIII answer. 

Substitute in eq. 3, using b%22%22=%22%22pi%2F4

1%22%22=%22%22e%5Ea%2Asin%28b%29

1%22%22=%22%22e%5Ea%2Asin%28pi%2F4%29

1%22%22=%22%22e%5Ea%2Aexpr%28sqrt%282%29%2F2%29

1%2F%28sqrt%282%29%2F2%29%22%22=%22%22e%5Ea

2%2Fsqrt%282%29%22%22=%22%22e%5Ea

expr%282%2Fsqrt%282%29%29%2Aexpr%28sqrt%282%29%2Fsqrt%282%29%29%22%22=%22%22e%5Ea

2sqrt%282%29%2F2%22%22=%22%22e%5Ea

sqrt%282%29%22%22=%22%22e%5Ea

Square both sides to eliminate the square root:

2%22%22=%22%22%28e%5Ea%29%5E2

2%22%22=%22%22e%5E%282a%29

Take ln of both sides:

ln%282%29%22%22=%22%22ln%28e%5E%282a%29%29

ln%282%29%22%22=%22%222a

ln%282%29%2F2%22%22=%22%22a

So the solution is

a%2Bbi%22%22=%22%22ln%282%29%2F2%2Bexpr%28pi%2F4%29%2Ai

Edwin