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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 ).
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"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