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Algebra Level pending

[ ln ( 1 + i ) ] \Re[ \; \ln (1+i) \; ]

Calculate the real part of the complex number ln ( 1 + i ) \ln (1+i ) correct to three decimal places.


The answer is 0.347.

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1 solution

Jason Simmons
Dec 28, 2015

Log ( z ) = ln z + i Arg ( z ) , z C \textrm{Log} (z) = \ln |z| +i \textrm{Arg} (z), \: z \in \mathbb{C}

Log ( 1 + i ) = ln 1 + i + i Arg ( 1 + i ) \textrm{Log} (1+i) = \ln |1+i| +i \textrm{Arg} (1+i)

Log ( 1 + i ) = ln 2 + i π 4 ln ( 1 + i ) = ln 2 + i π 4 \textrm{Log} (1+i) = \ln \sqrt{2} + \frac{i \pi}{4} \: \: \Rightarrow \: \: \ln (1+i) = \ln \sqrt{2} +\frac{i \pi}{4}

[ Log ( 1 + i ) ] = ln 2 0.347 \Re [ \; \textrm{Log} (1+i) \; ] = \ln \sqrt{2} \approx \boxed{0.347}

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