The Real one!

Algebra Level 4

Find the real part of ( 1 − i ) − i { \left( 1-i \right) }^{ -i }

Details and Assumptions :

i = − 1 i=\sqrt { -1 }

You can use scientific calculator to find the value of the real part in decimals.


The answer is 0.4288.

Mudit Bansal by mudit bansal , 25, India

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3 solutions

Let z = ( 1 − i ) − i z = (1-i)^{-i}

⇒ z = ( 2 e − π 4 i ) − i ⇒ ln ⁡ z = − i ( ln ⁡ 2 − π 4 i ) = − π 4 − ln ⁡ 2 i ⇒ z = e − π 4 e − ln ⁡ 2 i = e − π 4 [ cos ⁡ ( − ln ⁡ 2 ) + sin ⁡ ( − ln ⁡ 2 ) i ] ⇒ ℜ ( z ) = e − π 4 cos ⁡ ( − ln ⁡ 2 ) = 0.429 \begin{aligned} \Rightarrow z & = \left( \sqrt{2}e^{\frac{-\pi}{4}i} \right)^{-i} \\ \\ \Rightarrow \ln{z} & = -i \left( \ln{\sqrt{2}} - \frac{\pi}{4}i \right) \\ & = - \frac{\pi}{4} - \ln{\sqrt{2}}i \\ \\ \Rightarrow z & = e^{-\frac{\pi}{4}} e^{-\ln{\sqrt{2}}i} \\ & = e^{-\frac{\pi}{4}} \left[\cos{(-\ln{\sqrt{2}})}+\sin{(-\ln{\sqrt{2}})}i \right] \\ \\ \Rightarrow \Re {(z)} & = e^{-\frac{\pi}{4}} \cos{(-\ln{\sqrt{2}})} \\ & = \boxed{0.429} \end{aligned}

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In general 1 − i = 2 e i ( 7 π 4 + 2 n π ) \large 1 - i = \sqrt{2}e^{i(\frac{7\pi}{4} + 2n\pi)} , which for n = − 1 n = -1 is 2 e − i π 4 \large \sqrt{2}e^{-i\frac{\pi}{4}} .

The choice of n n would then appear to affect the final expression for z z as found using your method. WolframAlpha confirms your final answer, but I'm just wondering how we know we should choose n = − 1 n = -1 in order to start with the "correct" expression for 1 − i 1 - i ?

Brian Charlesworth - 5 years, 5 months ago
Mahek Mehta
Mar 7, 2015

Actually the author wants the absolute value of ( 1 − i ) − i (1-i)^{-i} ( 1 − i ) = 2 e − i π / 4 (1-i) = \sqrt{2}e^{-i\pi /4} , so ( 1 − i ) − i = 2 − i / 2 e − π / 4 (1-i)^{-i} = 2^{-i/2}e^{-\pi/4} Absolute value = e − p i / 4 e^{-pi/4} = 0.455

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No,you also need to convert 2^(-i)/2 in the form a+ib as it is not purely imaginary then you should report the real part.The right answer is 0.4288.It is a mistake by me in putting the answer.

mudit bansal - 6 years, 3 months ago
Pau Cantos
May 8, 2018

Notice that 1-i=exp(-π/4i+ln(2)/2) can bee expressed in the form exp(a+ib) as all the complex numbers. Then (1-i)^-1=exp(iln(2)/2-π/4)=exp(-π/4)exp(iln(2)/2) were exp(-π/4) is the absolute value and ln(2)/2 the argument. This its real part is exp(-π/4)cos(ln(2)/2)=0.428829...

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