For example if y is purely imaginary and x is purely real, then for z = 0, x and y must equal 0. If more complicated terms such as trigonometric ones are involved, how do you know that each respective term must be 0?
If a term $z = x+ y = 0$, how do you know when the additive terms, $x$ and $y$, must each be $0$?
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algebra-precalculus
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0Do you mean $z=x+iy$? – 2017-01-06
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0Well in general. – 2017-01-06
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0Then you should have that in the question... – 2017-01-06
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0What do you mean? It is the question – 2017-01-06
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0It says $z=x+y$, not $x+iy$. – 2017-01-06
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0y could be imaginary. – 2017-01-06
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0Could be? If not, then suppose $x=-y\ne0$. That is a solution. – 2017-01-06
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1Without extra conditions, you don't in general. – 2017-01-06
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0So is it only in the cases of imaginary numbers that x cannot = -y? – 2017-01-06
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0Also for example $\sqrt{x}+\sqrt{y}=0 \implies x=y=0$ – 2017-01-06
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0@Goldname Why `x cannot = -y` when `y` is imaginary? – 2017-01-06
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0You haven't properly set up your problem is what I'm trying to say. It isn't good notation to say $z=x+y$ when you mean $z=x+iy$. – 2017-01-06
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0@SimpleArt You're right, i forgot a condition. But your notation would be better only if z is strictly complex, but it might not be – 2017-01-06
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0@dxiv I edited q – 2017-01-06
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0If it is not complex, what is it? I'm sorry, but I thought we working with basic algebra here based on the tags. – 2017-01-06
5 Answers
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I'd think a more appropriate formulation is that $x$ and $y$ are part of the same algebraic structure $A$ (i.e. real, rational, complex, or integer numbers, or any of a number of other alternatives), and we have vectors (or something similar) $a,b,z$ from the same space, that may be multiplied by elements of $A$. Then, if $a$ and $b$ are linearly independent over $A$, that means that $z=xa+yb=0$ means $x=y=0$.
In the example in the comments, we have that $a$ is the complex number $1$, and $b$ is the complex number $i$. Complex numbers may be multiplied with real numbers, and $a$ and $b$ are linearly independent over $\Bbb R$. Thus $xa+yb=0$ means $x=y=0$ if $x$ and $y$ are limited to be real. The fact that real numbers can be thought of as special complex numbers means that this example may be expressed in a so-called "simplified" way that is easier to read and write, but less simple to see what is actually going on.
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Let $x,y$ be reals.
If $x+y=0$ with $xy\geq 0$ then $x=y=0$
If $x+iy=0$ then $x=y=0$.
If $ix+y=0$ then $x=y=0$.
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If $x+y=0$ then $x=-y$. But if $x$ is real and $y$ is pure imaginary, then $-y$ also pure imaginary and you cannot have $x=-y$. It doesn't matter if the variables represent more complicated functions - It's the fact that one is real and one is pure imaginary that prevents them from canceling each other.
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I think you are getting confused because of the following implication, if $x$ and $y$ are reals:
$$x+iy=0 \implies x=y=0$$
The way to prove this depends on how you defined the set of complex numbers, but you can see geometrically that if $x$ and $y$ are reals, then $x+iy$ can be mapped to the point $(x,y)$ of the plane. If $x+iy=0$, then the coordinates $x$ and $y$ are $0$.
In general, of course, it is not true ($1+(-1)=0$ but $1 \neq 0$). In fact, it is often a property of linear independance that allows us to infer such result.
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When $x$ and $y$ are real, you know this if you know that $x \geq 0$ and $y \geq 0$. For example:
- $a^2+b^2=0$. Squares are nonnegative, i.e. $a^2 \geq 0$ and $b^2 \geq 0$. So we have $a^2+b^2 \geq 0$ with equality if $a^2=0$ and $b^2=0$. In this case $x=a^2$, $y=b^2$.
- $\sqrt{a}+\sqrt{b}=0$. Again $\sqrt{a}$ and $\sqrt{b}$ are nonnegative, so here we have $\sqrt{a}=\sqrt{b}=0$.
- One with trigonometry: $(1+\sin(a))+(1-\cos(b))=0$ implies that $1+\sin(a)=0$ and $1-\cos(b)=0$ because $1+\sin(a) \geq 0$ and $1-\cos(b) \geq 0$.