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If we have a complex number number $a + ib = 0$, then we can write as $a + bi = 0 + 0\cdot i$. Equating corresponding terms, we get $a = b = 0$.

What is this? Where am I going wrong ?

Watson
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shraman shraman
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    Um... isn't that correct? – Frank Oct 18 '16 at 17:43
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    Are $a,b$ supposed to be real? If so, then if $a+bi=0$ with $a,b$ real, then $a=b=0$. But it is not true if $a,b$ are complex. For example $a=1,b=i$ has $a+bi=0$. – Thomas Andrews Oct 18 '16 at 17:46
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    What do you mean where are you going wrong? You're not going wrong anywhere. 0 = 0 + 0i is a true statement. If a+bi = 0 then a= b = 0 is a true statement. What are you having a problem with? – fleablood Oct 18 '16 at 18:04
  • You are correct. $z=a+bi=0$ iff $Re(z)=a=0$ and $Im(z)=b=0$. – Addy Oct 18 '16 at 18:07
  • @fleablood The answers below all assume that $a, b \in \mathbb{R}$ but I don't see that spelled out in question itself. – dxiv Oct 19 '16 at 01:33
  • True. But it's a convention that when say "let a +bi is a complex number" that we are talking of a complex number with real part a and imaginary part b. True, this wasn't explicitly stated but... well, again, what exactly is the question? If a = -bi there are a variety of answers including a=b=0. – fleablood Oct 19 '16 at 02:37
  • I wonder if the issue is re (z) and im (z) are different concepts and we are getting the odd looking case of a complex number where re (z)=z itself. As well as im(z)=z itself... which is fine if z=0.... – fleablood Oct 19 '16 at 02:46

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I don't quite understand your confusion, it is correct if a complex number $z = a+ib= 0$ (with $a,b\in\mathbb{R}$), then both its real part and its imaginary part are zero, i.e. $\text{Re}(z)=a=0$, and $\text{Im} (z) = b = 0$.

Eff
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  • This can be proved in this way: if $b \neq 0$, then $i$ would be equal to $-a/b$, which is a real number, so we can't have $(-a/b)^2 = -1$. Therefore $b=0$, so that $a+bi=a+0i=a=0$. – Watson Oct 18 '16 at 17:49
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A complex number, Z, has the form $Z = x+iy$ where $x$ and $y$ are real numbers and $i$ is the imaginary unit postulated such that $i$$^2$ = $-1$. Quantity $x$ is the real part of $Z$ and $y$ is the imaginary part of $Z$. Please note that $y$ is a Real quantity.

Therefore, A Real Number is thus a complex number with $0$ Imaginary Part.

A complex number with $0$ Real Part is a pure imaginary number.

The number which is both Real and Pure imaginary at the same time is $0$.

It is, therefore, trivial to write $0$ as $0$ + $i0$ which hardly serves any meaningful purpose except for the sake of brevity. With this background, you are going wrong NOWHERE.

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"If we have a complex number number a+ib=0, then we can write as a+bi=0+0⋅i. Equating corresponding terms, we get a=b=0"

Yes, that is correct. $0 = 0 + 0i. Re(0) = 0. Im(0) = 0$. If $z = a + bi = 0 + 0i$ then $a = Re(0) = Re(0 + 0i) = 0; b = Im(0) = Im(0+0i) = 0;$ and $a = b = 0$.

I suppose part of the confusion is if $x \in \mathbb R \subset \mathbb C$ then $x = x + 0i$ and $Re(x) = x$ and $Im(x) = 0$ and if $x = a + bi$ then $a =x$ and $b = 0$. But ... that shouldn't be confusing. As a real number $x$ is "purely" real its imaginary component must be $0$.

.

"What is this?"

Um... A sentence????

"Where am I going wrong ? "

Nowhere.

fleablood
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This amounts to say that $1$ and $i$ are linearly independent, i.e. the only solution of

$$\lambda\cdot1+\mu\cdot i=0$$ is $$\lambda=\mu=0.$$

We can prove it by multiplying the equation by $i$ to form a system

$$\begin{cases}\ \ \ \lambda+\mu\, i=0\\-\mu+\lambda\,i=0.\end{cases}$$

It has a nonzero determinant ($2i$), which justifies the claim.