For z- 0- define log z - log-z- - iarg z where - - argz -i-e- argz stands for the principal argument of z- log-i equals -

The correct option is C π i/2 log( i) =log(e^ iπ2) ...(using Euler's form) = iπ2 ...

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Byju's Answer

Standard XII

Mathematics

nth Term of HP

For z≠ 0, d...

Question

For $z \neq 0$ , define $log z = log | z | + i (a r g z)$ where $- π < a r g (z) \leq π$
i.e. $a r g (z)$ stands for the principal argument of $z$ .
$log (- i)$ equals :

π i / 2

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π i

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- π i / 2

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Solution

The correct option is C $- π i / 2$
$l o g (- i)$
$= l o g (e^{- i \frac{π}{2}})$ ...(using Euler's form)
$= - i \frac{π}{2}$

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Similar questions

For

z \neq 0

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log z = log | z | + i (a r g z)

where

- π < a r g (z) \leq π

i.e.

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For z≠0, define logz=log|z|+i(argz) where −π<arg(z)≤πi.e. arg(z) stands for the principal argument of z.log(−i) equals :

The correct option is C −πi/2log(−i)=log(e−iπ2) ...(using Euler's form)=−iπ2

For $z \neq 0$ , define $log z = log | z | + i (a r g z)$ where $- π < a r g (z) \leq π$
i.e. $a r g (z)$ stands for the principal argument of $z$ .
$log (- i)$ equals :

The correct option is C $- π i / 2$
$l o g (- i)$
$= l o g (e^{- i \frac{π}{2}})$ ...(using Euler's form)
$= - i \frac{π}{2}$