How to take an absolute value or modulus of z?












0














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
























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










  • @Crazyformaths yes
    – LenaPark
    Dec 8 at 9:37










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
















0














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
























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










  • @Crazyformaths yes
    – LenaPark
    Dec 8 at 9:37










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














0












0








0







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















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 at 18:45









Shaun

8,507113580




8,507113580










asked Dec 8 at 9:20









LenaPark

83




83












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










  • @Crazyformaths yes
    – LenaPark
    Dec 8 at 9:37










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


















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










  • @Crazyformaths yes
    – LenaPark
    Dec 8 at 9:37










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
















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




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












@Crazyformaths yes
– LenaPark
Dec 8 at 9:37




@Crazyformaths yes
– LenaPark
Dec 8 at 9:37












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




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










2 Answers
2






active

oldest

votes


















1














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






share|cite|improve this answer





















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



















1














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





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














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






    share|cite|improve this answer





















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
















    1














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






    share|cite|improve this answer





















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














    1












    1








    1






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






    share|cite|improve this answer












    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 at 9:24









    Shaun

    8,507113580




    8,507113580












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


















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
















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




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











    1














    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


























      1














      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
























        1












        1








        1






        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












        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 at 9:42









        egreg

        177k1484200




        177k1484200






























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