For any two complex numbers $z_1$ and $z_2$, prove that $\operatorname{Re}(z_1 z_2) = \operatorname{Re} z_1\, \operatorname{Re} z_2 - \operatorname{Im} z_1\, \operatorname{Im} z_2$.

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> [!TLDR]
> The real part of the product of two complex numbers equals the product of their real parts minus the product of their imaginary parts — this follows directly from expanding $(a+ib)(c+id)$ using the distributive law and the fact that $i^2 = -1$.

The identity you're asked to prove is one of the most fundamental in complex arithmetic. It tells us that when you multiply two complex numbers, the real part of the result behaves exactly like the dot product of two vectors in the plane — which is no accident, because complex numbers *are* vectors in disguise.

Let’s see why this works.

1. **Write both numbers in standard form.**  
   Any complex number can be written as $z = x + iy$, where $x = \operatorname{Re} z$ and $y = \operatorname{Im} z$. So let  
   $$z_1 = a + ib, \quad z_2 = c + id$$  
   with $a, b, c, d$ all real numbers. Here $a = \operatorname{Re} z_1$, $b = \operatorname{Im} z_1$, $c = \operatorname{Re} z_2$, $d = \operatorname{Im} z_2$.

2. **Multiply them out — carefully.**  
   $$z_1 z_2 = (a + ib)(c + id)$$  
   Use the distributive law (FOIL):  
   $$= a c + a(id) + ib(c) + ib(id)$$  
   $$= ac + i\,ad + i\,bc + i^2 bd$$

3. **Replace $i^2$ with $-1$.**  
   That’s the single most important rule: $i^2 = -1$. So the last term becomes $-bd$:  
   $$z_1 z_2 = ac + i(ad + bc) - bd$$  
   Group the real and imaginary parts:  
   $$z_1 z_2 = (ac - bd) + i(ad + bc)$$

4. **Read off the real part.**  
   The real part of $z_1 z_2$ is the term without $i$:  
   $$\operatorname{Re}(z_1 z_2) = ac - bd$$  
   But $a = \operatorname{Re} z_1$, $c = \operatorname{Re} z_2$, $b = \operatorname{Im} z_1$, $d = \operatorname{Im} z_2$. Substituting gives exactly  
   $$\operatorname{Re}(z_1 z_2) = \operatorname{Re} z_1 \cdot \operatorname{Re} z_2 - \operatorname{Im} z_1 \cdot \operatorname{Im} z_2$$

> [!WARNING]
> A common mistake is to think $\operatorname{Re}(z_1 z_2) = (\operatorname{Re} z_1)(\operatorname{Re} z_2)$ — that would be true only if the imaginar

Q.For any two complex numbers $z_1$ and $z_2$ , prove that $\operatorname{Re}(z_1 z_2) = \operatorname{Re} z_1\, \operatorname{Re} z_2 - \operatorname{Im} z_1\, \operatorname{Im} z_2$ .

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Q.For any two complex numbers z1z_1z1​ and z2z_2z2​, prove that Re⁡(z1z2)=Re⁡z1 Re⁡z2−Im⁡z1 Im⁡z2\operatorname{Re}(z_1 z_2) = \operatorname{Re} z_1\, \operatorname{Re} z_2 - \operatorname{Im} z_1\, \operatorname{Im} z_2Re(z1​z2​)=Rez1​Rez2​−Imz1​Imz2​.

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Q.For any two complex numbers $z_1$ and $z_2$ , prove that $\operatorname{Re}(z_1 z_2) = \operatorname{Re} z_1\, \operatorname{Re} z_2 - \operatorname{Im} z_1\, \operatorname{Im} z_2$ .

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