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(B) Let $z_1 = a + ib$ and $z_2 = c + id$,where $a, b, c, d \in \mathbb{R}$.Since $z_1 + z_2 = (a + c) + i(b + d)$ is real,the imaginary part must be

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$z_1 = \bar{z}_2$

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(B) Let $z_1 = a + ib$ and $z_2 = c + id$,where $a, b, c, d \in \mathbb{R}$.Since $z_1 + z_2 = (a + c) + i(b + d)$ is real,the imaginary part must be zero: $b + d = 0$,which implies $d = -b$.Since $z_1 z_2 = (a + ib)(c + id) = (ac - bd) + i(ad + bc)$ is real,the imaginary part must be zero: $ad + bc = 0$.Substituting $d = -b$ into the equation $ad + bc = 0$,we get $a(-b) + bc = 0$,which simplifies to $b(c - a) = 0$.Case $1$: If $b = 0$,then $d = 0$. Thus $z_1 = a$ and $z_2 = c$,which are real numbers. In this case,$z_1 = \bar{z}_1$ and $z_2 = \bar{z}_2$. This does not uniquely determine the relationship between $z_1$ and $z_2$ unless they are equal or conjugate.Case $2$: If $b 
eq 0$,then $c = a$. Since $d = -b$,we have $z_2 = a - ib = \bar{z}_1$,which implies $z_1 = \bar{z}_2$.Thus,the condition $z_1 = \bar{z}_2$ satisfies the given requirements.

Let $z_1$ and $z_2$ be two complex numbers such that $z_1 + z_2$ and $z_1 z_2$ are both real. Then:

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