If $f$ and $g$ both satisfy Cauchy Riemann equations at point $z$, prove $f+g$ and $fg$ satisfy the Cauchy...












0














So if $f=u+iv$, and $g=m+in$. Trying to prove $fg$ follows the Cauchy Riemann equations: I got $fg=(mu-nv)+i(nu+mv)$, letting $a=mu-nv$ and letting $b=nu+mv$, I differentiated $a$ and $b$, by $x$ and $y$ but I can't equate the differentials so I don't know what to do.






complex-analysis differential-equations







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  • Hint: you don't need to differentiate anything at all. You already know that $frac{partial u}{partial x} = frac{partial v}{partial y}$, and that $frac{partial m}{partial x} = frac{partial n}{partial y}$, and you know that (partial) differentiation is a linear map.
    – user3482749
    Nov 18 at 22:53










  • I don't understand what you mean by (partial) differentiation is a linear map.
    – M. Calculator
    Nov 18 at 22:56










  • The set of all differentiable functions $mathbb{C}tomathbb{C}$ is a vector space over $mathbb{C}$. The maps $frac{partial}{partial x}$ and $frac{partial}{partial y}$ are linear maps from that vector space to $mathbb{C}$.
    – user3482749
    Nov 19 at 15:32
















0














So if $f=u+iv$, and $g=m+in$. Trying to prove $fg$ follows the Cauchy Riemann equations: I got $fg=(mu-nv)+i(nu+mv)$, letting $a=mu-nv$ and letting $b=nu+mv$, I differentiated $a$ and $b$, by $x$ and $y$ but I can't equate the differentials so I don't know what to do.






complex-analysis differential-equations







share|cite|improve this question
























  • Hint: you don't need to differentiate anything at all. You already know that $frac{partial u}{partial x} = frac{partial v}{partial y}$, and that $frac{partial m}{partial x} = frac{partial n}{partial y}$, and you know that (partial) differentiation is a linear map.
    – user3482749
    Nov 18 at 22:53










  • I don't understand what you mean by (partial) differentiation is a linear map.
    – M. Calculator
    Nov 18 at 22:56










  • The set of all differentiable functions $mathbb{C}tomathbb{C}$ is a vector space over $mathbb{C}$. The maps $frac{partial}{partial x}$ and $frac{partial}{partial y}$ are linear maps from that vector space to $mathbb{C}$.
    – user3482749
    Nov 19 at 15:32














0












0








0







So if $f=u+iv$, and $g=m+in$. Trying to prove $fg$ follows the Cauchy Riemann equations: I got $fg=(mu-nv)+i(nu+mv)$, letting $a=mu-nv$ and letting $b=nu+mv$, I differentiated $a$ and $b$, by $x$ and $y$ but I can't equate the differentials so I don't know what to do.






complex-analysis differential-equations







share|cite|improve this question















So if $f=u+iv$, and $g=m+in$. Trying to prove $fg$ follows the Cauchy Riemann equations: I got $fg=(mu-nv)+i(nu+mv)$, letting $a=mu-nv$ and letting $b=nu+mv$, I differentiated $a$ and $b$, by $x$ and $y$ but I can't equate the differentials so I don't know what to do.






complex-analysis differential-equations




complex-analysis differential-equations






share|cite|improve this question















share|cite|improve this question













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edited Nov 18 at 23:20









Joey Kilpatrick

1,181422




1,181422










asked Nov 18 at 22:49









M. Calculator

376




376












  • Hint: you don't need to differentiate anything at all. You already know that $frac{partial u}{partial x} = frac{partial v}{partial y}$, and that $frac{partial m}{partial x} = frac{partial n}{partial y}$, and you know that (partial) differentiation is a linear map.
    – user3482749
    Nov 18 at 22:53










  • I don't understand what you mean by (partial) differentiation is a linear map.
    – M. Calculator
    Nov 18 at 22:56










  • The set of all differentiable functions $mathbb{C}tomathbb{C}$ is a vector space over $mathbb{C}$. The maps $frac{partial}{partial x}$ and $frac{partial}{partial y}$ are linear maps from that vector space to $mathbb{C}$.
    – user3482749
    Nov 19 at 15:32


















  • Hint: you don't need to differentiate anything at all. You already know that $frac{partial u}{partial x} = frac{partial v}{partial y}$, and that $frac{partial m}{partial x} = frac{partial n}{partial y}$, and you know that (partial) differentiation is a linear map.
    – user3482749
    Nov 18 at 22:53










  • I don't understand what you mean by (partial) differentiation is a linear map.
    – M. Calculator
    Nov 18 at 22:56










  • The set of all differentiable functions $mathbb{C}tomathbb{C}$ is a vector space over $mathbb{C}$. The maps $frac{partial}{partial x}$ and $frac{partial}{partial y}$ are linear maps from that vector space to $mathbb{C}$.
    – user3482749
    Nov 19 at 15:32
















Hint: you don't need to differentiate anything at all. You already know that $frac{partial u}{partial x} = frac{partial v}{partial y}$, and that $frac{partial m}{partial x} = frac{partial n}{partial y}$, and you know that (partial) differentiation is a linear map.
– user3482749
Nov 18 at 22:53




Hint: you don't need to differentiate anything at all. You already know that $frac{partial u}{partial x} = frac{partial v}{partial y}$, and that $frac{partial m}{partial x} = frac{partial n}{partial y}$, and you know that (partial) differentiation is a linear map.
– user3482749
Nov 18 at 22:53












I don't understand what you mean by (partial) differentiation is a linear map.
– M. Calculator
Nov 18 at 22:56




I don't understand what you mean by (partial) differentiation is a linear map.
– M. Calculator
Nov 18 at 22:56












The set of all differentiable functions $mathbb{C}tomathbb{C}$ is a vector space over $mathbb{C}$. The maps $frac{partial}{partial x}$ and $frac{partial}{partial y}$ are linear maps from that vector space to $mathbb{C}$.
– user3482749
Nov 19 at 15:32




The set of all differentiable functions $mathbb{C}tomathbb{C}$ is a vector space over $mathbb{C}$. The maps $frac{partial}{partial x}$ and $frac{partial}{partial y}$ are linear maps from that vector space to $mathbb{C}$.
– user3482749
Nov 19 at 15:32










1 Answer
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Note thatbegin{align}frac{partial(mu-nv)}{partial x}&=frac{partial m}{partial x}u+frac{partial u}{partial x}m-frac{partial n}{partial x}v-frac{partial v}{partial x}n\&=frac{partial n}{partial y}u+frac{partial v}{partial y}m+frac{partial m}{partial y}v+frac{partial u}{partial y}n\&=frac{partial n}{partial y}u+frac{partial u}{partial y}n+frac{partial v}{partial y}m+frac{partial m}{partial y}v\&=frac{partial(nu+mv)}{partial y}.end{align}Can you take it from here?






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  • I don't understand how the first line is equal to the second line, as the first line is differentiated with respect x an the third line is differentiated with respect to y. Also, the first line of the 3rd and 4th terms are negative whereas there are no negatives on the 2nd line.
    – M. Calculator
    Nov 18 at 23:15












  • I am assuming that $f$ and $g$ satisfy the Cauchy-RIemann equations. Therefore, $frac{partial m}{partial x}=frac{partial n}{partial y}$, $frac{partial n}{partial x}=-frac{partial m}{partial y}$, $frac{partial u}{partial x}=frac{partial v}{partial y}$, and $frac{partial u}{partial y}=-frac{partial v}{partial x}$.
    – José Carlos Santos
    Nov 18 at 23:22










  • I see, thank you for that. Do I use the same process for f+g?
    – M. Calculator
    Nov 18 at 23:42










  • Sure. It's even simpler in that case.
    – José Carlos Santos
    Nov 18 at 23:44










  • Thank you so much.
    – M. Calculator
    Nov 18 at 23:47











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Note thatbegin{align}frac{partial(mu-nv)}{partial x}&=frac{partial m}{partial x}u+frac{partial u}{partial x}m-frac{partial n}{partial x}v-frac{partial v}{partial x}n\&=frac{partial n}{partial y}u+frac{partial v}{partial y}m+frac{partial m}{partial y}v+frac{partial u}{partial y}n\&=frac{partial n}{partial y}u+frac{partial u}{partial y}n+frac{partial v}{partial y}m+frac{partial m}{partial y}v\&=frac{partial(nu+mv)}{partial y}.end{align}Can you take it from here?






share|cite|improve this answer





















  • I don't understand how the first line is equal to the second line, as the first line is differentiated with respect x an the third line is differentiated with respect to y. Also, the first line of the 3rd and 4th terms are negative whereas there are no negatives on the 2nd line.
    – M. Calculator
    Nov 18 at 23:15












  • I am assuming that $f$ and $g$ satisfy the Cauchy-RIemann equations. Therefore, $frac{partial m}{partial x}=frac{partial n}{partial y}$, $frac{partial n}{partial x}=-frac{partial m}{partial y}$, $frac{partial u}{partial x}=frac{partial v}{partial y}$, and $frac{partial u}{partial y}=-frac{partial v}{partial x}$.
    – José Carlos Santos
    Nov 18 at 23:22










  • I see, thank you for that. Do I use the same process for f+g?
    – M. Calculator
    Nov 18 at 23:42










  • Sure. It's even simpler in that case.
    – José Carlos Santos
    Nov 18 at 23:44










  • Thank you so much.
    – M. Calculator
    Nov 18 at 23:47
















2














Note thatbegin{align}frac{partial(mu-nv)}{partial x}&=frac{partial m}{partial x}u+frac{partial u}{partial x}m-frac{partial n}{partial x}v-frac{partial v}{partial x}n\&=frac{partial n}{partial y}u+frac{partial v}{partial y}m+frac{partial m}{partial y}v+frac{partial u}{partial y}n\&=frac{partial n}{partial y}u+frac{partial u}{partial y}n+frac{partial v}{partial y}m+frac{partial m}{partial y}v\&=frac{partial(nu+mv)}{partial y}.end{align}Can you take it from here?






share|cite|improve this answer





















  • I don't understand how the first line is equal to the second line, as the first line is differentiated with respect x an the third line is differentiated with respect to y. Also, the first line of the 3rd and 4th terms are negative whereas there are no negatives on the 2nd line.
    – M. Calculator
    Nov 18 at 23:15












  • I am assuming that $f$ and $g$ satisfy the Cauchy-RIemann equations. Therefore, $frac{partial m}{partial x}=frac{partial n}{partial y}$, $frac{partial n}{partial x}=-frac{partial m}{partial y}$, $frac{partial u}{partial x}=frac{partial v}{partial y}$, and $frac{partial u}{partial y}=-frac{partial v}{partial x}$.
    – José Carlos Santos
    Nov 18 at 23:22










  • I see, thank you for that. Do I use the same process for f+g?
    – M. Calculator
    Nov 18 at 23:42










  • Sure. It's even simpler in that case.
    – José Carlos Santos
    Nov 18 at 23:44










  • Thank you so much.
    – M. Calculator
    Nov 18 at 23:47














2












2








2






Note thatbegin{align}frac{partial(mu-nv)}{partial x}&=frac{partial m}{partial x}u+frac{partial u}{partial x}m-frac{partial n}{partial x}v-frac{partial v}{partial x}n\&=frac{partial n}{partial y}u+frac{partial v}{partial y}m+frac{partial m}{partial y}v+frac{partial u}{partial y}n\&=frac{partial n}{partial y}u+frac{partial u}{partial y}n+frac{partial v}{partial y}m+frac{partial m}{partial y}v\&=frac{partial(nu+mv)}{partial y}.end{align}Can you take it from here?






share|cite|improve this answer












Note thatbegin{align}frac{partial(mu-nv)}{partial x}&=frac{partial m}{partial x}u+frac{partial u}{partial x}m-frac{partial n}{partial x}v-frac{partial v}{partial x}n\&=frac{partial n}{partial y}u+frac{partial v}{partial y}m+frac{partial m}{partial y}v+frac{partial u}{partial y}n\&=frac{partial n}{partial y}u+frac{partial u}{partial y}n+frac{partial v}{partial y}m+frac{partial m}{partial y}v\&=frac{partial(nu+mv)}{partial y}.end{align}Can you take it from here?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 18 at 23:11









José Carlos Santos

150k22121221




150k22121221












  • I don't understand how the first line is equal to the second line, as the first line is differentiated with respect x an the third line is differentiated with respect to y. Also, the first line of the 3rd and 4th terms are negative whereas there are no negatives on the 2nd line.
    – M. Calculator
    Nov 18 at 23:15












  • I am assuming that $f$ and $g$ satisfy the Cauchy-RIemann equations. Therefore, $frac{partial m}{partial x}=frac{partial n}{partial y}$, $frac{partial n}{partial x}=-frac{partial m}{partial y}$, $frac{partial u}{partial x}=frac{partial v}{partial y}$, and $frac{partial u}{partial y}=-frac{partial v}{partial x}$.
    – José Carlos Santos
    Nov 18 at 23:22










  • I see, thank you for that. Do I use the same process for f+g?
    – M. Calculator
    Nov 18 at 23:42










  • Sure. It's even simpler in that case.
    – José Carlos Santos
    Nov 18 at 23:44










  • Thank you so much.
    – M. Calculator
    Nov 18 at 23:47


















  • I don't understand how the first line is equal to the second line, as the first line is differentiated with respect x an the third line is differentiated with respect to y. Also, the first line of the 3rd and 4th terms are negative whereas there are no negatives on the 2nd line.
    – M. Calculator
    Nov 18 at 23:15












  • I am assuming that $f$ and $g$ satisfy the Cauchy-RIemann equations. Therefore, $frac{partial m}{partial x}=frac{partial n}{partial y}$, $frac{partial n}{partial x}=-frac{partial m}{partial y}$, $frac{partial u}{partial x}=frac{partial v}{partial y}$, and $frac{partial u}{partial y}=-frac{partial v}{partial x}$.
    – José Carlos Santos
    Nov 18 at 23:22










  • I see, thank you for that. Do I use the same process for f+g?
    – M. Calculator
    Nov 18 at 23:42










  • Sure. It's even simpler in that case.
    – José Carlos Santos
    Nov 18 at 23:44










  • Thank you so much.
    – M. Calculator
    Nov 18 at 23:47
















I don't understand how the first line is equal to the second line, as the first line is differentiated with respect x an the third line is differentiated with respect to y. Also, the first line of the 3rd and 4th terms are negative whereas there are no negatives on the 2nd line.
– M. Calculator
Nov 18 at 23:15






I don't understand how the first line is equal to the second line, as the first line is differentiated with respect x an the third line is differentiated with respect to y. Also, the first line of the 3rd and 4th terms are negative whereas there are no negatives on the 2nd line.
– M. Calculator
Nov 18 at 23:15














I am assuming that $f$ and $g$ satisfy the Cauchy-RIemann equations. Therefore, $frac{partial m}{partial x}=frac{partial n}{partial y}$, $frac{partial n}{partial x}=-frac{partial m}{partial y}$, $frac{partial u}{partial x}=frac{partial v}{partial y}$, and $frac{partial u}{partial y}=-frac{partial v}{partial x}$.
– José Carlos Santos
Nov 18 at 23:22




I am assuming that $f$ and $g$ satisfy the Cauchy-RIemann equations. Therefore, $frac{partial m}{partial x}=frac{partial n}{partial y}$, $frac{partial n}{partial x}=-frac{partial m}{partial y}$, $frac{partial u}{partial x}=frac{partial v}{partial y}$, and $frac{partial u}{partial y}=-frac{partial v}{partial x}$.
– José Carlos Santos
Nov 18 at 23:22












I see, thank you for that. Do I use the same process for f+g?
– M. Calculator
Nov 18 at 23:42




I see, thank you for that. Do I use the same process for f+g?
– M. Calculator
Nov 18 at 23:42












Sure. It's even simpler in that case.
– José Carlos Santos
Nov 18 at 23:44




Sure. It's even simpler in that case.
– José Carlos Santos
Nov 18 at 23:44












Thank you so much.
– M. Calculator
Nov 18 at 23:47




Thank you so much.
– M. Calculator
Nov 18 at 23:47


















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