$$\int z \,dz=\dfrac{z^2}{2}$$ $$z=a+bi\implies \int (a+bi) \,dz=\int z \, dz$$ $$a(a+bi)+bi(a+bi)=(a+bi)^2=\dfrac{z^2}{2}=\dfrac{(a+bi)^2}{2}$$ $$ 2(a+bi)^2=(a+bi)^2$$ Assume $z \neq 0$: $$1=2$$ Where is the fallacy?
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$\begingroup$ $2(a+bi)^2\not=(a+bi)^2$, I believe... $\endgroup$– Conor O'BrienCommented Dec 9, 2014 at 4:20
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3$\begingroup$ Where are you getting your third line from? $\endgroup$– Akiva WeinbergerCommented Dec 9, 2014 at 4:20
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$\begingroup$ How did you conclude that $a(a+bi)+bi(a+bi)$ has something to do with evaluating the integrals? $\endgroup$– Jonas MeyerCommented Dec 9, 2014 at 4:22
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2$\begingroup$ @MathNoob: You are treating $z$ as constant and not constant simultaneously. $\endgroup$– Jonas MeyerCommented Dec 9, 2014 at 4:25
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1$\begingroup$ The first line is wrong to begin with. You have omitted the arbitrary additive constant. Then the conclusion in the second line is wrong. You can't replace a dummy variable with a constant because then there is nothing to integrate; that's likely the error you are looking for. $\endgroup$– MPWCommented Dec 9, 2014 at 4:32
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1 Answer
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In your third line you write $z^2=\displaystyle\frac{z^2}2$ out of the blue. This only happens when $z=0$, so you cannot conclude that $1=2$.
Ok, now I got what you were trying to do. When you "evaluate the integrals" you are taking $a$ and $b$ as constants. That makes no sense if $a+ib=z$ and $z$ is your integration variable.
More concretely, $$a=\frac{z+\bar z}2,\ b=\frac{z-\bar z}{2i}.$$. You cannot treat them as constants.
answered Dec 9, 2014 at 4:23
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$\begingroup$ Evaluating the integral $\int (a+bi) dz$ gives $a(z)+bi(z)=z(a+bi)=z^2$ Evaluating the second gives z^2/2. $\endgroup$– TeocCommented Dec 9, 2014 at 4:24
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4$\begingroup$ No it doesn't. If $a$ and $b$ do not depend on $z$, then you cannot write $z=a+ib$. $\endgroup$ Commented Dec 9, 2014 at 4:26