Why is $2Re\{c_n e^{\frac{j2n\pi t}{T}} \}$ equal to $2Re\{c_n \} cos{\frac{2n\pi t}{T}} - 2Im \{ c_n \} sin{ \frac{2n\pi t}{T}} $?

Question

Why is $2Re\{c_n e^{\frac{j2n\pi t}{T}} \}$ equal to $2Re\{c_n \} cos{\frac{2n\pi t}{T}} - 2Im \{ c_n \} sin{ \frac{2n\pi t}{T}} $ in the Fourier series?

See images below:

$C_n$

Derivation

Answer 1

If you have a complex number $z$ you can write it as $\Re(z)+i\Im(z)$ where $\Re$ and $\Im$ are the real and imaginary functions and return purely real values.

So consider the product of two complex numbers $z_1$ and $z_2$:

$z_1\cdot z_2=(\Re(z_1)+i\Im(z_1))(\Re(z_2)+i\Im(z_2))$

$z_1\cdot z_2=(\Re(z_1)\Re(z_2)-\Im(z_1\Im(z_2))+i(\Re(z_1)\Im(z_2)+\Re(z_2)\Im(z_1))$

So $\Re(z_1\cdot z_2)=\Re(z_1)\Re(z_2)-\Im(z_1)\Im(z_2)$

Apply this to your situation: $c_n e^\frac{j2n\pi t}{T}$

$z_1=c_n$ and $z_2=e^\frac{j2n\pi t}{T}$

The real and imaginary functions applied to $e^{i\theta}$ gives $\cos\theta$ and $\sin\theta$ respectively. This leads to your answer.

Answer 2

Elementary, consider to complex numbers a and b, $$ a = a_1+i*a_2\ and\ b = b_1+i*b_2 $$ $$ Re(a*b) = Re((a_1+i a_2)*(b_1+i*b_2)) = Re((a_1*b_1-a_2*b_2)+i*(a_1*b_2+a_2*b_1)) = a_1*b_1-a_2*b_2 = Re(a)*Re(b)-Im(a)*Im(b) $$ S0 if you consider $C_n$ as complex number $$ Re(C_n*e^{j2n\pi t/T}) = Re(C_n)*Re(e^{j2n\pi t/T}) - Im(C_n)*Im(e^{j2n\pi t/T}) = Re(C_n)*cos(j2n\pi t/T) - Im(C_n)*sin(j2n\pi t/T) $$