How to take an absolute value or modulus of z?












0












$begingroup$


let's assume the :



$$z=frac{e^{(-jc)}}{(a+jb)}$$
I would like to take the absolute value of z.
I started with multiplication z with $frac{(a-jb)}{(a-jb)}$ and got:
$$absfrac{e^{(-jc)}}{(a+jb)}=frac{e^{(-jc)}(a-jb)}{(a^2+b^2)}$$
Then:
$$frac{e^{(-jc)}(a-jb)*e^{(jc)}}{(a^2-b^2)e^{(jc)}}=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$$
Should I repeat the calculations or I made something wrong?



edit 1:



$(a-jb)(a+jb)=a^2+b^2$



edit 2:



$ abs(e^{(-jc)})=1$



$ abs(z)=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$



what should i do with $(a-jb)$? Or will be it $(a^2+b^2)$?






complex-analysis complex-numbers modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you using 'j' as a symbol for iota(i.e. $sqrt{-1}$)?
    $endgroup$
    – Martund
    Dec 8 '18 at 9:23










  • $begingroup$
    @Crazyformaths yes
    $endgroup$
    – LenaPark
    Dec 8 '18 at 9:37










  • $begingroup$
    Multiplication with $a-jb$ gives $a^2+b^2$ in the denominator.
    $endgroup$
    – Martund
    Dec 8 '18 at 9:39
















0












$begingroup$


let's assume the :



$$z=frac{e^{(-jc)}}{(a+jb)}$$
I would like to take the absolute value of z.
I started with multiplication z with $frac{(a-jb)}{(a-jb)}$ and got:
$$absfrac{e^{(-jc)}}{(a+jb)}=frac{e^{(-jc)}(a-jb)}{(a^2+b^2)}$$
Then:
$$frac{e^{(-jc)}(a-jb)*e^{(jc)}}{(a^2-b^2)e^{(jc)}}=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$$
Should I repeat the calculations or I made something wrong?



edit 1:



$(a-jb)(a+jb)=a^2+b^2$



edit 2:



$ abs(e^{(-jc)})=1$



$ abs(z)=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$



what should i do with $(a-jb)$? Or will be it $(a^2+b^2)$?






complex-analysis complex-numbers modules







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you using 'j' as a symbol for iota(i.e. $sqrt{-1}$)?
    $endgroup$
    – Martund
    Dec 8 '18 at 9:23










  • $begingroup$
    @Crazyformaths yes
    $endgroup$
    – LenaPark
    Dec 8 '18 at 9:37










  • $begingroup$
    Multiplication with $a-jb$ gives $a^2+b^2$ in the denominator.
    $endgroup$
    – Martund
    Dec 8 '18 at 9:39














0












0








0





$begingroup$


let's assume the :



$$z=frac{e^{(-jc)}}{(a+jb)}$$
I would like to take the absolute value of z.
I started with multiplication z with $frac{(a-jb)}{(a-jb)}$ and got:
$$absfrac{e^{(-jc)}}{(a+jb)}=frac{e^{(-jc)}(a-jb)}{(a^2+b^2)}$$
Then:
$$frac{e^{(-jc)}(a-jb)*e^{(jc)}}{(a^2-b^2)e^{(jc)}}=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$$
Should I repeat the calculations or I made something wrong?



edit 1:



$(a-jb)(a+jb)=a^2+b^2$



edit 2:



$ abs(e^{(-jc)})=1$



$ abs(z)=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$



what should i do with $(a-jb)$? Or will be it $(a^2+b^2)$?






complex-analysis complex-numbers modules







share|cite|improve this question











$endgroup$




let's assume the :



$$z=frac{e^{(-jc)}}{(a+jb)}$$
I would like to take the absolute value of z.
I started with multiplication z with $frac{(a-jb)}{(a-jb)}$ and got:
$$absfrac{e^{(-jc)}}{(a+jb)}=frac{e^{(-jc)}(a-jb)}{(a^2+b^2)}$$
Then:
$$frac{e^{(-jc)}(a-jb)*e^{(jc)}}{(a^2-b^2)e^{(jc)}}=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$$
Should I repeat the calculations or I made something wrong?



edit 1:



$(a-jb)(a+jb)=a^2+b^2$



edit 2:



$ abs(e^{(-jc)})=1$



$ abs(z)=frac{(a-jb)}{(a^2+b^2)e^{(jc)}}$



what should i do with $(a-jb)$? Or will be it $(a^2+b^2)$?






complex-analysis complex-numbers modules




complex-analysis complex-numbers modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 18:45









Shaun

9,083113683




9,083113683










asked Dec 8 '18 at 9:20









LenaParkLenaPark

83




83












  • $begingroup$
    Are you using 'j' as a symbol for iota(i.e. $sqrt{-1}$)?
    $endgroup$
    – Martund
    Dec 8 '18 at 9:23










  • $begingroup$
    @Crazyformaths yes
    $endgroup$
    – LenaPark
    Dec 8 '18 at 9:37










  • $begingroup$
    Multiplication with $a-jb$ gives $a^2+b^2$ in the denominator.
    $endgroup$
    – Martund
    Dec 8 '18 at 9:39


















  • $begingroup$
    Are you using 'j' as a symbol for iota(i.e. $sqrt{-1}$)?
    $endgroup$
    – Martund
    Dec 8 '18 at 9:23










  • $begingroup$
    @Crazyformaths yes
    $endgroup$
    – LenaPark
    Dec 8 '18 at 9:37










  • $begingroup$
    Multiplication with $a-jb$ gives $a^2+b^2$ in the denominator.
    $endgroup$
    – Martund
    Dec 8 '18 at 9:39
















$begingroup$
Are you using 'j' as a symbol for iota(i.e. $sqrt{-1}$)?
$endgroup$
– Martund
Dec 8 '18 at 9:23




$begingroup$
Are you using 'j' as a symbol for iota(i.e. $sqrt{-1}$)?
$endgroup$
– Martund
Dec 8 '18 at 9:23












$begingroup$
@Crazyformaths yes
$endgroup$
– LenaPark
Dec 8 '18 at 9:37




$begingroup$
@Crazyformaths yes
$endgroup$
– LenaPark
Dec 8 '18 at 9:37












$begingroup$
Multiplication with $a-jb$ gives $a^2+b^2$ in the denominator.
$endgroup$
– Martund
Dec 8 '18 at 9:39




$begingroup$
Multiplication with $a-jb$ gives $a^2+b^2$ in the denominator.
$endgroup$
– Martund
Dec 8 '18 at 9:39










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: If $w=re^{jtheta}$ for $r,thetainBbb R$, then $|w|=|r|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is assuming $j^2=-1$.
    $endgroup$
    – Shaun
    Dec 8 '18 at 9:25



















1












$begingroup$

Recall that
$$
left|frac{z_1}{z_2}right|=frac{|z_1|}{|z_2|}
$$
If $c$ is real, then $|e^{-jc}|=|cos c+jsin c|=1$. So you have
$$
left|frac{e^{-jc}}{a+jb}right|=frac{1}{|a+jb|}=frac{1}{sqrt{a^2+b^2}}
$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint: If $w=re^{jtheta}$ for $r,thetainBbb R$, then $|w|=|r|$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is assuming $j^2=-1$.
      $endgroup$
      – Shaun
      Dec 8 '18 at 9:25
















    1












    $begingroup$

    Hint: If $w=re^{jtheta}$ for $r,thetainBbb R$, then $|w|=|r|$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is assuming $j^2=-1$.
      $endgroup$
      – Shaun
      Dec 8 '18 at 9:25














    1












    1








    1





    $begingroup$

    Hint: If $w=re^{jtheta}$ for $r,thetainBbb R$, then $|w|=|r|$.






    share|cite|improve this answer









    $endgroup$



    Hint: If $w=re^{jtheta}$ for $r,thetainBbb R$, then $|w|=|r|$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 8 '18 at 9:24









    ShaunShaun

    9,083113683




    9,083113683












    • $begingroup$
      This is assuming $j^2=-1$.
      $endgroup$
      – Shaun
      Dec 8 '18 at 9:25


















    • $begingroup$
      This is assuming $j^2=-1$.
      $endgroup$
      – Shaun
      Dec 8 '18 at 9:25
















    $begingroup$
    This is assuming $j^2=-1$.
    $endgroup$
    – Shaun
    Dec 8 '18 at 9:25




    $begingroup$
    This is assuming $j^2=-1$.
    $endgroup$
    – Shaun
    Dec 8 '18 at 9:25











    1












    $begingroup$

    Recall that
    $$
    left|frac{z_1}{z_2}right|=frac{|z_1|}{|z_2|}
    $$
    If $c$ is real, then $|e^{-jc}|=|cos c+jsin c|=1$. So you have
    $$
    left|frac{e^{-jc}}{a+jb}right|=frac{1}{|a+jb|}=frac{1}{sqrt{a^2+b^2}}
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Recall that
      $$
      left|frac{z_1}{z_2}right|=frac{|z_1|}{|z_2|}
      $$
      If $c$ is real, then $|e^{-jc}|=|cos c+jsin c|=1$. So you have
      $$
      left|frac{e^{-jc}}{a+jb}right|=frac{1}{|a+jb|}=frac{1}{sqrt{a^2+b^2}}
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Recall that
        $$
        left|frac{z_1}{z_2}right|=frac{|z_1|}{|z_2|}
        $$
        If $c$ is real, then $|e^{-jc}|=|cos c+jsin c|=1$. So you have
        $$
        left|frac{e^{-jc}}{a+jb}right|=frac{1}{|a+jb|}=frac{1}{sqrt{a^2+b^2}}
        $$






        share|cite|improve this answer









        $endgroup$



        Recall that
        $$
        left|frac{z_1}{z_2}right|=frac{|z_1|}{|z_2|}
        $$
        If $c$ is real, then $|e^{-jc}|=|cos c+jsin c|=1$. So you have
        $$
        left|frac{e^{-jc}}{a+jb}right|=frac{1}{|a+jb|}=frac{1}{sqrt{a^2+b^2}}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 9:42









        egregegreg

        181k1485203




        181k1485203






























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