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Thread: Complex numbers

  1. #1
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    Complex numbers

    Complex numbers-test2.png
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  2. #2
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    Re: logarithmic functions and their graphs

    Quote Originally Posted by retro View Post
    Click image for larger version. 

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    Although I belong to a group of analysts who object to using the radical signs with complex numbers here is some help.
    $\large{(-3\mathcal{i})(3\mathcal{i})=9}$

    $\large{(2+\mathcal{i})(2-\mathcal{i})=5}$
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    Re: logarithmic functions and their graphs

    How would u represent it without using radical signs
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    Re: logarithmic functions and their graphs

    Quote Originally Posted by retro View Post
    How would u represent it without using radical signs
    Use the 1/2 power: $\displaystyle \left(\frac{(-3i)(3i)}{(2+ i)(2- i)}\right)^{1/2}$


    What I know: common logarithms use base-10.

    What I think I know: a logarithm is an exponent.


    I don't know how to parlay my limited knowledge into a solution, please help
    What was wrong with doing exactly what you were told to do?
    "Choose two numbers". Okay, I choose 3 an 7. "Take their common logarithms" log(3)= 0.4771 and log(7)= 0.8451. "Add the logarithms" log(3)+ log(7)= 0.4771+ 0.8451= 1.3222. "Take the logarithm of their product". 3*7= 21 and log(21)= 1.3222. Hmmm! Now choose three numbers and try it again!

    In the future post two different questions in different threads. People who have already responded to the first question might not look at the thread again and not see the new question.
    Last edited by HallsofIvy; May 6th 2019 at 01:13 PM.
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    Re: logarithmic functions and their graphs

    Quote Originally Posted by retro View Post
    How would u represent it without using radical signs
    If you are still asking about $\sqrt{\dfrac{(-3i)(3i)}{(2+i)(2-i)}}$ then

    Quote Originally Posted by Plato View Post
    Although I belong to a group of analysts who object to using the radical signs with complex numbers here is some help.
    $\large{(-3\mathcal{i})(3\mathcal{i})=9}$
    $\large{(2+\mathcal{i})(2-\mathcal{i})=5}$
    So that we have
    $ \begin{align*}\sqrt{\dfrac{(-3i)(3i)}{(2+i)(2-i)}}&=\sqrt{\dfrac{9}{5}} \\&=\dfrac{3}{\sqrt 5}\\&=\dfrac{3\sqrt 5}{5} \end{align*}$
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