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Math Help - real and imaginary

  1. #1
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    real and imaginary

    Hi, can someone please help me write  Log [(1+i)^4] in terms of its real and imaginary parts?
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    write

    1+i = \sqrt{2}\cdot (e^{i\frac{\pi}{4}})

    Look at it for some time
    ...do you need more help than that .?
    feel free to ask.
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  3. #3
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    omigosh i am so lost in this subject

    coz i thought that i had to use  Log z = ln|z| + i Arg(z)

    well this is what i did, but like yeahh i really dont know what im doing
     ln|1+i|^4 + i Arg(1+i)^4
     4 ln|1+i|  + i Arg(1+i)^4

    umm please help me more sorry i didnt get how you got that answer
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  4. #4
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    Convert to polars.

    x + iy = r(\cos{\theta} + i\sin{\theta}) = r\,e^{i\theta}


    You have 1 + i, which is in the first quadrant.


    r = \sqrt{1^2 + 1^2} = \sqrt{2}.


    \theta = \arctan{\frac{1}{1}} = \frac{\pi}{4}.


    Therefore 1 + i = \sqrt{2}\left(\cos{\frac{\pi}{4}} + i\sin{\frac{\pi}{4}}\right) = \sqrt{2}\,e^{\frac{\pi}{4}i}.



    So (1 + i)^4 = (\sqrt{2}\,e^{\frac{\pi}{4}i})^4

     = 4\,e^{\pi i}

     = -4.



    Can you see that

    \log{[(1 + i)^4]} = \log{(-4)}

     = \ln{|-4|} + i\arg{(-4)}

     = \ln{4} + \pi i

     = 2\ln{2} + \pi i.


    Now it is written in terms of its real and imaginary parts.
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