Check the correct sign $pm$ in $( Ve^{-j(wt+A)})/(Ie^{-j(wt+B)})=(V/I)cdot[cos(A-B) pm jsin(A-B)]$












-1












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If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$




I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$






proof-verification complex-numbers







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  • $begingroup$
    If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
    $endgroup$
    – Shubham Johri
    Dec 16 '18 at 10:30
















-1












$begingroup$



If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$




I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$






proof-verification complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
    $endgroup$
    – Shubham Johri
    Dec 16 '18 at 10:30














-1












-1








-1





$begingroup$



If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$




I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$






proof-verification complex-numbers







share|cite|improve this question











$endgroup$





If $v = Ve^{-j(wt+A)}, i= Ie^{-j(wt+B)}$ then show that the impedance $z = v/i$ is given by $$Z = (V/I)cdot[cos(A-B) + jsin(A-B)]$$




I get confused because when I used the correct rules, it always comes to $$(V/I)cdot[cos(A-B) - jsin(A-B)]$$ instead because of the rule that $e^{-j(theta)}=cos(theta) - sin(theta)$






proof-verification complex-numbers




proof-verification complex-numbers






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 16 '18 at 11:22









Did

248k23223460




248k23223460










asked Dec 16 '18 at 10:12









Mvb1Mvb1

1




1












  • $begingroup$
    If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
    $endgroup$
    – Shubham Johri
    Dec 16 '18 at 10:30


















  • $begingroup$
    If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
    $endgroup$
    – Shubham Johri
    Dec 16 '18 at 10:30
















$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30




$begingroup$
If $j=sqrt{-1}$, your answer is correct (although you might want to check the rule $e^{-j(theta)}=cos(theta) - sin(theta)$ again)
$endgroup$
– Shubham Johri
Dec 16 '18 at 10:30










1 Answer
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$begingroup$

Well,



$$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
by Euler's formula, then because cosine is even and sine is odd,
$$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$



You got it






share|cite|improve this answer









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    $begingroup$

    Well,



    $$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
    by Euler's formula, then because cosine is even and sine is odd,
    $$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$



    You got it






    share|cite|improve this answer









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      0












      $begingroup$

      Well,



      $$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
      by Euler's formula, then because cosine is even and sine is odd,
      $$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$



      You got it






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Well,



        $$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
        by Euler's formula, then because cosine is even and sine is odd,
        $$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$



        You got it






        share|cite|improve this answer









        $endgroup$



        Well,



        $$ frac{v}{i} = frac{V}{I}e^{-j(A-B)} = frac{V}{I} [cos(-(A-B))+jsin(-(A-B))]$$
        by Euler's formula, then because cosine is even and sine is odd,
        $$ frac{v}{i} = frac{V}{I} [cos(A-B) -j sin(A-B)].$$



        You got it







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 16 '18 at 10:37









        kevinkayakskevinkayaks

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