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Please explain how the solve the image problem please explain fully clearly......!..............

sudhanva gv , 6 Years ago
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Samyak Jain

Last Activity: 6 Years ago

1 + i = \sqrt{2} (1/\sqrt{2} + i.1/\sqrt{2})  =  \sqrt{2}(cos\pi/4 + i sin\pi/4)  =  \sqrt{2}.e^(i.\pi/4)   [Euler’s form of complex number]
\therefore z  =  log2 (1 + i)  =  log2 (\sqrt{2}.e^(i.\pi/4))
and z(bar)  =  log2 (\sqrt{2}.e^(– i.\pi/4))
z + z(bar)  =  log2 (\sqrt{2}.e^(i.\pi/4))  +  log2 (\sqrt{2}.e^(– i.\pi/4))
                =  log2 (\sqrt{2}.e^(i.\pi/4).\sqrt{2}.e^(– i.\pi/4))  =  log2 (2)
                =  1                                                  [\because logc a + logc b = logc ab and logaa = 1]
z – z(bar) = log2 (\sqrt{2}.e^(– i.\pi/4)) – log2 (\sqrt{2}.e^(i.\pi/4))  =  log2 (\sqrt{2}.e^(i.\pi/4) / (\sqrt{2}.e^(– i.\pi/4))
               = log2 (e^(i\pi/4 + i\pi/4))  =  log2 (e^(i\pi/2))         [\because logc a – logc b = logc a/b]
               =  loge (e^(i\pi/2)) / loge 2               using change of base law of logarithm.
               = (i\pi/2) / ln2
               = i \pi/2.ln2  =  i.\pi/ln22  =  i.\pi/ln4
\therefore z + z(bar) + i(z – z(bar))  =  1 + i(i.\pi/ln4)  =  1 + i2.\pi/ln4 = 1 – \pi/ln4   [i2 = – 1]
                                        = (ln4 – \pi) / ln4.

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