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?
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.