prove-Log-z-1-z-2-Log-z-1-Log-z-2-if-pi-lt-Argz-1-Argz-2-lt-pi-hwo-can-solve-this-

Question Number 132826 by mohammad17 last updated on 16/Feb/21

p r o v e L o g (z_{1} z_{2}) = L o g (z_{1}) + L o g (z_{2})

i f - π < {A r g z}_{1} + {A r g z}_{2} < π

h w o c a n s o l v e t h i s

Commented by guyyy last updated on 20/Feb/21

Answered by MJS_new last updated on 17/Feb/21

z_{1} = r e^{i α} \land z_{2} = s e^{i β}

\log z_{1} z_{2} = \log r s e^{i (α + β)} = \log r + \log s + i (α + β)

\log z_{1} + \log z_{2} = \log r e^{i α} + \log s e^{i β} =

= \log r + i α + \log s + i β = \log r + \log s + i (α + β)

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