Log(tanπ/4+ix/2)=?

1 Answer
Feb 15, 2018

# \ #

# \qquad log( tan(\pi/4) + i x/2 ) \ = \ log ( \sqrt{ 4 + x^2 }/2 ) + i tan^{-1}( x/2 ). #

Explanation:

# \ #

# log( tan(\pi/4) + i x/2 ) \ = \ log( 1 + i x/2 ) #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \ log( 1/2 \cdot [ 2 + i x ] ) #

# \qquad \qquad \qquad = \ log [ 1/2 \cdot \sqrt{ 4 + x^2 } \cdot ( 2 / \sqrt{ 4 + x^2 } + i x / \sqrt{ 4 + x^2 } ) ] #

# \qquad \qquad \qquad = \ log [ \sqrt{ 4 + x^2 }/2 \cdot ( 2 / \sqrt{ 4 + x^2 } + i x / \sqrt{ 4 + x^2 } ) ] #

# \qquad \qquad \qquad = \ \ log ( \sqrt{ 4 + x^2 }/2 ) + log( 2 / \sqrt{ 4 + x^2 } + i x / \sqrt{ 4 + x^2 } ) #

# \qquad \qquad \qquad = \ \ log ( \sqrt{ 4 + x^2 }/2 ) + log[ cos( \theta ) + i sin( \theta ) ]; #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad "where" \quad \theta \ = \ tan^{-1}( x/2 ) #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad "(see footnote)" #

# \qquad \qquad \qquad = \ \ log ( \sqrt{ 4 + x^2 }/2 ) + log[ e^{ i \theta } ] #

# \qquad \qquad \qquad = \ \ log ( \sqrt{ 4 + x^2 }/2 ) + i \theta #

# \qquad \qquad \qquad = \ \ log ( \sqrt{ 4 + x^2 }/2 ) + i tan^{-1}( x/2 ); #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad "since" \quad \theta \ = \ tan^{-1}( x/2 ). #

# "So we have our result:" #

# \qquad log( tan(\pi/4) + i x/2 ) \ = \ log ( \sqrt{ 4 + x^2 }/2 ) + i tan^{-1}( x/2 ). #

# \ #

# "Footnote: making a right triangle with one leg" = 2, #
# "other leg" \ = x,\ "and angle" \ \ \theta = \ "the angle adjacent to the" #
# "leg of measure 2, we see immediately:" #

# cos( \theta ) = 2 / \sqrt{ 4 + x^2 }, \qquad sin( \theta ) = x / \sqrt{ 4 + x^2 }, \qquad tan( \theta ) = x / 2. #

# "By the last equation here, we have at once:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \theta = tan^{ -1 }( x / 2 ). #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \square #