what-the-value-of-log-10-1-in-complex-number-

Question Number 116725 by bemath last updated on 06/Oct/20

what the value of \log_{10} (- 1) in

complex number

Commented by bemath last updated on 06/Oct/20

thank you prof

Commented by MJS_new last updated on 06/Oct/20

usually we take the principal value

z = r e^{i θ} with r \in R^{+} and - π ⩽ θ < π []

\Rightarrow principal value of \ln z = \ln r + i θ

and \log_{z_{2}} z_{1} = \frac{\ln z_{1}}{\ln z_{2}} = \frac{\ln r_{1} e^{i θ_{1}}}{\ln r_{2} e^{i θ_{2}}} = \frac{\ln r_{1} + i θ_{1}}{\ln r_{2} + i θ_{2}} =

= \frac{θ_{1} θ_{2} + \ln r_{1} \ln r_{2}}{\ln^{2} r_{2} + θ_{2}^{2}} + i \frac{θ_{1} \ln r_{2} - θ_{2} \ln r_{1}}{\ln^{2} r_{2} + θ_{2}^{2}}

with r_{1}, r_{2} > 0

Commented by MJS_new last updated on 06/Oct/20

btw also \ln e^{r} = r is “ {only}^{″} the principal

value : e^{r} = e^{r} e^{2 n π i} \Rightarrow \ln e^{r} = r + 2 n π i but

who would like to use this ?

Answered by MJS_new last updated on 06/Oct/20

\log_{10} (- 1) = \frac{\ln (- 1)}{\ln 10} = \frac{\ln e^{i π}}{\ln 10} = \frac{i π}{\ln 10}

Commented by bemath last updated on 06/Oct/20

sir if \log_{10} (0) = ?

Commented by bemath last updated on 06/Oct/20

- 1 = \cos π + i . \sin π, sir

Answered by TANMAY PANACEA last updated on 06/Oct/20

t r y i n g

L o g (a + i b)

a = r c o s θ b = r s i n θ

L o g ({r e}^{i θ}) = L o g ({r e}^{i (2 k π + θ)}) = l o g r + i (2 k π + θ)

L o g (a + i b) = l o g (\sqrt{a^{2} + b^{2}}) + i {2 k π + {t a n}^{- 1} (\frac{b}{a})}

L o g (a + i b) = \frac{1}{2} l o g (a^{2} + b^{2}) + i (2 k π + {t a n}^{- 1} (\frac{b}{a}))

{l o g}_{10} p \times x = {l o g}_{e} p

\frac{l o g p}{l o g 10} \times x = \frac{l o g p}{l o g e} \to x = {l o g}_{e} 10

{l o g}_{10} (- 1) = {l o g}_{e} (- 1) \times {l o g}_{10} e

{l o g}_{10} (- 1)

= {l o g}_{e} (- 1) \times {l o g}_{10} e

= [\frac{1}{2} l n (1 + 0^{2}) + i (2 k π + {t a n}^{- 1} (\frac{- 1}{0})] {l o g}_{10} e

= [0 + i (2 k π - \frac{π}{2})] {l o g}_{10} e

Commented by bemath last updated on 06/Oct/20

how about \log_{10} (0) sir

Commented by Rio Michael last updated on 06/Oct/20

now am sure \log_{10} (0) should be undefined

since \ln (0) is undefined .

Commented by Dwaipayan Shikari last updated on 06/Oct/20

Y e s!

Commented by bemath last updated on 06/Oct/20

ok . thank you all

Answered by Rio Michael last updated on 06/Oct/20

i know \ln (- 1) = π i

\log (- 1) = \frac{\ln (- 1)}{\ln 10} = \frac{π i}{\ln 10}

Answered by Dwaipayan Shikari last updated on 06/Oct/20

{l o g}_{10} (1 - x) = \frac{l o g (1 - x)}{l o g 10} = - \frac{1}{l o g 10} (x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots .)

X = 1

- \frac{1}{l o g 10} (1 + \frac{1}{2} + \frac{1}{3} + \dots .) \to - \infty

Commented by Dwaipayan Shikari last updated on 06/Oct/20

B a s i c a l l y 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots . i s u n d e f i n e d

B u t g e n e r a l l y t a k e i t a s d i v e r g e s \dots