I'm currently learning about complex numbers, and I was wondering: isn't $33i$ the same as $0+33i$ and therefore has a real and imaginary part. Can $33i$ be considered as real and complex?

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    $\begingroup$ Real: no. Complex: yes. $\endgroup$ – quasi Oct 9 '17 at 8:01
  • $\begingroup$ $33i$ cannot be real because $33i$ is a product of two numbers $33, i$ for which $33 = 33$ is real and $i = \sqrt{-1}$ which is not real. However we can re-write that $33 = 33 + 0i$ and show that every real number is complex, but no complex numbers are real. From this we can also show that $33i = 0 + 33i$ so it is complex, hence not real. We have a set $\mathbb{R}$ denoting the set of all real numbers, and a set $\mathbb{C}$ denoting the set of all complex numbers. $\mathbb{R} \subseteq \mathbb{C}$ but $\mathbb{C} \not\subseteq \mathbb{R}$ $\endgroup$ – Feeds Oct 9 '17 at 8:08
  • $\begingroup$ You may find this helpful if you do not understand the difference between an imaginary number and a complex number since a good argument is that $i = \sqrt{-1}$ is imaginary, but also equals $0 + 1i$ so $i$ is also complex $\longrightarrow$ math.stackexchange.com/questions/304207/… $\endgroup$ – Feeds Oct 9 '17 at 8:16
  • $\begingroup$ @user477343 You're not wrong, but I'd be careful with the "product of real and not real" argument, as 0i would often be considered real. $\endgroup$ – Mark S. Oct 9 '17 at 10:42
  • $\begingroup$ Oh yes, rehashing on what Mark said, $0i$ is real because $0i = 0$ and $0$ is real. Sorry about that, thanks @MarkS. $\endgroup$ – Feeds Oct 9 '17 at 20:46

No. Real numbers have no imaginary part.

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    $\begingroup$ "Real numbers have an imaginary part equal to zero". $\endgroup$ – Glougloubarbaki Oct 9 '17 at 10:09
  • $\begingroup$ @Glougloubarbaki no, that doesn't really work. It would only be true if you constructed the set of complex numbers first and then defined the reals as a subset thereof, but that's backwards (such a deductive working process doesn' scale); what's actually done is, you construct the reals in some suitable way, define the complex numbers in terms of them, and then declare the canonical embedding $\mathbb{R} \to \mathbb{C}$ which maps the reals to a subfield of $\mathbb{C}$. But e.g. $5 : \mathbb{C}$ is not the same thing and $5 : \mathbb{R}$, it's just equal under that implicit embedding. $\endgroup$ – leftaroundabout Oct 9 '17 at 12:59
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    $\begingroup$ @leftaroundabout while that is technically correct, and even something good to keep in mind from time to time, that sort of thinking is ridiculously impractical. Do you also distinguish between $2$ and $\frac{2}{1}$, or use frequently $0=\emptyset$ ? $\endgroup$ – Glougloubarbaki Oct 9 '17 at 13:49
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    $\begingroup$ @Glougloubarbaki as a matter of fact, I do generally distinguish between $2:\mathbb{Z}$ and $\frac21:\mathbb{Q}$ and $2.0:\mathbb{R}$ and $2.0 + 2.0i : \mathbb{C}$. I still tend to write all of them simply as $2$, but that I understood as a polymorphic number literal which can represent different things depending on the context. This can be made rigorous in a Hindley-Milner type theory framework. Usually, that mindset comes down to the same as pretending “$\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\subset\mathbb{C}$”. $\endgroup$ – leftaroundabout Oct 9 '17 at 14:06
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    $\begingroup$ @leftaroundabout ok then. but I'm not sure if it is best to emphasize that mindset to undergrads discovering complex numbers in a first calculus course... $\endgroup$ – Glougloubarbaki Oct 9 '17 at 15:01

All complex numbers have a real part and an imaginary part.

Every complex number $z$ has a unique representation in the form $$z = a + bi$$ where $a,b$ are real numbers.

By definition, the real part of $z$ is $a$, and the imaginary part of $z$ is $b$.

If $b=0$, then $z=a + 0i = a$, so $z$ is real.

Conversely, if $z$ is real, say $z = a$, then one representation of $z$ is $z = a + 0i$, hence by the uniqueness property, if $z = a + bi$, we must have $b=0$.

Therefore, $z$ is real if and only if $b=0$.

In words, a complex number is real if and only if its imaginary part is equal to zero.

Thus, $33i = 0 + 33i\;$has $a=0,\;$and $b=33,\;$hence, since $b \ne 0$, $33i$ is complex but not real.

That's based on the definitions.

But here's another way to see that $33i$ is not real . . .

If $33i$ was real, then dividing by $33$, it would follow that $i$ is real.

By definition, we have $i^2 = -1$.

Now suppose, for the sake of argument, that $i$ is real.

Then either $i > 0$, $i < 0$, or $i=0$.

If $i > 0$, then $i^2$ would be positive (since "a positive times a positive is a positive").

If $i < 0$, then $i^2$ would be positive (since "a negative times a negative is a positive").

If $i = 0$, then of course, $i^2=0$.

In all three cases, we get $i^2 \ge 0$, contradiction, since $i^2 = -1$.

Thus, $i$ is not real, and hence, neither is $33i$.


A complex number is a number that can be expressed in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit (which satisfies the equation $i^2 = −1$).

$33i=0+33i=a+b i$ where $a=0,b=33$. Ofc $0,33$ are real numbers and hence your number is a complex number.


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