Difference between revisions of "Logarithm of product is a sum of logarithms"

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==Theorem==
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Let $z_1,z_2 \in \mathbb{C}$ such that $-\pi < \mathrm{arg}(z_1) + \mathrm{arg}(z_2) \leq \pi$. Then the following formula holds:
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$$\log(z_1z_2) = \log(z_1)+\log(z_2),$$
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where $\log$ denotes the [[logarithm]].
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==Proof==
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==References==
 
==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Logarithm (multivalued) of product is a sum of logarithms (multivalued)|next=}}: 4.1.7
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* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Logarithm (multivalued) of product is a sum of logarithms (multivalued)|next=Logarithm (multivalued) of a quotient is a difference of logarithms (multivalued)}}: 4.1.7

Revision as of 06:16, 4 June 2016

Theorem

Let $z_1,z_2 \in \mathbb{C}$ such that $-\pi < \mathrm{arg}(z_1) + \mathrm{arg}(z_2) \leq \pi$. Then the following formula holds: $$\log(z_1z_2) = \log(z_1)+\log(z_2),$$ where $\log$ denotes the logarithm.

Proof

References