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Verify Euler's formula for $e^{ix}$ by considering $\frac{dz}{dx}$ where $z=r(\cos x+i\sin x)$

I tried taking the derivative of z but could not get to Euler's from there.

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HINT:

$$\frac{dz}{dx}=r(-\sin x+i\cos x)=ri\cdot z$$

$$\implies\frac{dz}z=ri\cdot dx$$

Now integrate either sides and set $x=0$ to identify the value of arbitrary integral constant

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$$\frac{d(e^{ix})}{dx}=ie^{ix}$$

$$\frac{d(\cos x+i\sin x)}{dx}=-\sin x+i\cos x=i(\cos x+i\sin x)$$

Thus, both functions $\;f(x)=e^{ix}\;,\;\;g(x)=\cos x+i\sin x\;$ fulfill the condition

$$f'=if\;,\;\;g'=ig\implies \left(\log f\right)'=i=\left(\log g\right)'\implies$$

$$\log f=\log g+C\;,\;\;C=\text{constant}\implies$$

$$f=Kg\;,\;\;K=\text{constant}$$

Now evaluate both functions at $\;x=0\;$ to get the desired equality.

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let $f(x)=\cos(x)+i\sin(x)$ then we get $f'(x)=-\sin(x)+i\cos(x)=if(x)$ therefor $\ln(f(x))=ix+C$ from $f(0)=1$ we get $C=0$ thus we get $e^{ix}=\cos(x)+i\sin(x)$

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This is just an alternative approach. Consider asking the question what is the value of cos(ix) in polar form. That is find A(x) and B(x) such that $$\cos(ix)=A+iB$$ Differentiating the equality a couple of times $$-i\sin(ix)=A'+iB'$$ $$\cos(ix) = A" +iB"$$ Thus $$A"=A$$ $$B"=B$$ The solution for A or B is $C_1e^x+C_2e^{-x}$. For arbitrary $C_1$ and $C_2$. Something very-clear at the time the equation was developed. Both the question and the solution of the differential $A"=A$ where known at the time the equation was developed.

$$\cos(ix)=C_1e^x+C_2e^{-x} +i[C_3e^x+C_4e^{-x}]$$

$$\cos(i(-ix))=\cos(x)=C_1e^{-ix}+C_2e^{ix} +i[C_3e^{-ix}+C_4e^{ix}]$$ $$\cos(x)=C_1e^{-ix}+C_2e^{ix} +i[C_3e^{-ix}+C_4e^{ix}]$$ $$-sin(x)=C_1(-i)e^{-ix}+C_2(i)e^{ix} +i[C_3(-i)e^{-ix}+C_4(i)e^{ix}]$$ Applying BCs $\cos(0)=1$ and $\sin(0)= 0$ one finds that $C_1=C_2=1/2$ and $C_3=C_4=0$ fit the conditions.

Hence: $$\cos(x)=\frac{e^{-ix}+e^{ix}}{2}$$ $$-\sin(x)=\frac{-ie^{-ix}+ie^{ix}}{2}$$ $$-i\sin(x)=\frac{e^{-ix}-e^{ix}}{2}$$ Thus $$\cos(x)+i\sin(x)=e^{ix}$$

More on the BCs.

$$cos(0)=[C_1+C_2] +i[C_3+C_4]=1$$ $$[C_1+C_2]=1, [C_3+C_4]=0$$ $$-sin(0)=i[C_2-C_1] +[C_3-C_4]=0$$ $$[C_2-C_1]=0, [C_3-C_4]=0$$

Thus $C_1=C_2=1/2$ and $C_3=C_4=0$ fit the conditions. (I cannot demonstrate uniqueness for solution for $C_n$. In so far I accept the uniqueness because the same procedure apply to $cos(ix)$, $sin(ix)$, and $e^{ix}$ yield the same result: Euler's equation.)

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  • $\begingroup$ How did you get $$-i\sin(x)=A'+iB'$$ when differentiating ? $\endgroup$ – Vivek Kaushik Feb 17 '17 at 2:57
  • $\begingroup$ @ Vivek Kaushik, corrected. Please continue to point out errors. $\endgroup$ – 926reals Feb 17 '17 at 3:13
  • $\begingroup$ I don't understand your solution to the differential equations. $A''=A, B"=B.$ You say the solution to either $A$ or $B$ is $$Ae^{x}+Be^{-x},$$ shouldn't be $C_1e^{x}+C_2 e^{-x}$ for arbitrary constants $C_1,C_2$ ? $\endgroup$ – Vivek Kaushik Feb 17 '17 at 3:13

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