The real part of 1- i - i is -RPET 1999- A- e - pi - 4cos1-2log 2B- e - pi - 4sin1-2log 2C- e p i - 4cos1-2log 2D- e - pi - 4sin1-2log 2

Let z = (1 i)^ i. Taking log on both sides, ⇒ log z = i log(1 i) = i log √(2)(cosπ/4 i sinπ/4 ) = i log (√(2e)^ iπ/4) = i [1/2 log 2 + log e^ iπ/4 ] ...

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Standard XII

Mathematics

Euler's Representation

The real part...

Question

The real part of $(1 - i)^{- i}$ is
[RPET 1999]

e^{- p i / 4} c o s (\frac{1}{2} l o g 2)

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e^{- p i / 4} s i n (\frac{1}{2} l o g 2)

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e^{p i / 4} c o s (\frac{1}{2} l o g 2)

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e^{- p i / 4} s i n (\frac{1}{2} l o g 2)

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Solution

The correct option is A $e^{- p i / 4} c o s (\frac{1}{2} l o g 2)$
Let z = $(1 - i)^{- i} .$ Taking log on both sides,
$\Rightarrow l o g z = - i l o g (1 - i) = - i l o g \sqrt{2} (c o s \frac{π}{4} - i s i n \frac{π}{4})$
$= - i l o g ({\sqrt{2 e}}^{- i π / 4}) = - i [\frac{1}{2} l o g 2 + l o g e^{- i π / 4}]$
$i [\frac{1}{2} l o g 2 - \frac{i π}{4}] = - \frac{i}{2} l o g 2 - \frac{π}{4}$
$\Rightarrow z = e^{- π / 4} e^{- i / 2 l o g 2} .$ Taking real part only,
$\Rightarrow R e (z) = e^{π / 4} c o s (\frac{1}{2} l o g 2) .$

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Find the value of i. $i^{i}$ is _______.

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The real part of (1−i)−i is [RPET 1999]

The real part of $(1 - i)^{- i}$ is
[RPET 1999]