Let $z_1,z_2,\cdots,z_n$ be some complex numbers. If $z_1z_2\cdots z_n$ is real and negative, is it true that $r(z_1)r(z_2)\cdots r(z_n)$ is also negative? Here $r(z)$ represents the real part of $z$.
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For an strictly positive example, let $\omega=\frac{1+i\sqrt{3}}2$. Then $\omega^3=-1$, but $\Re(\omega)^3=\frac{1}{8}$. |
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Consider $z_1 =z_2 = i$. Then $z_1z_2 = -1$, but the product of the real parts $0\cdot 0 = 0 $ is not negative. |
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We have $i\cdot i=-1$ and yet $\mathfrak{R}(i)\cdot \mathfrak{R}(i)=0\cdot 0 = 0$. |
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The real part of $i^2$ is $-1$, but the product of the real part of $i$ is 0. |
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Multiplication of complex numbers is, geometrically, a rotation. Here is what that means. The angle between a complex number and the positive real number line is called the argument. When you multiply two complex numbers together, the resulting product's argument is the sum of the arguments of the two numbers. The argument of a positive, real number is zero. The argument of a negative real number is 180 degrees (or $\pi$ radians, since more commonly, the argument will be expressed in radians). The argument of $i$ is $\pi/2$ radians, and so when we multiply $i$ by $i$, the product must have an argument of $\pi$ radians, which is consistent with it being $-1$. Thus, any set of complex numbers whose arguments add up to $\pi$ (or to $\pi + 2\pi k$, where $k$ is an integer) will produce a negative real number if they are multiplied together. All of the numbers in such a set can have a real part greater than zero. It's possible for all the numbers in such a set to have a real part less than zero, or any mixture. |
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