I'm confused as to how to approach this problem. We are learning about the real and imaginary parts of a complex number.

$z=x+iy$ $$ \operatorname{Re} [(2+i)z] $$

I multiplied out $2+i$ with $x+iy$ and then took the real parts to get $2x-y$. But it doesn't feel like it's right. That seems too simple.


Ok so the resounding response seems to be that I did it right. However how do you all explain this website? It defines $\operatorname{Re} z$ and $\operatorname{Im} z$ a bit differently.

The real and imaginary parts are given by:

$$ \operatorname{Re} z = \frac {z + \bar z} 2 \\ \operatorname{Im} z = \frac {z - \bar z} {2i} $$

  • 1
    $\begingroup$ What you did is correct, instead. $\endgroup$
    – Crostul
    Commented Jan 18, 2016 at 17:50
  • 2
    $\begingroup$ Complex numbers are simple. $\endgroup$
    – Did
    Commented Jan 18, 2016 at 17:53
  • 1
    $\begingroup$ @Did They are simple and yet complex. $\endgroup$
    – Mark Viola
    Commented Jan 18, 2016 at 18:05
  • 1
    $\begingroup$ Why did you expect it to be difficult? Okay, this is usually a bad idea but let's look as how a ... less adequate... student could have gone wrong. "(2 + i)z; the 2 is real and the i is not so the answer is 2z" or "(2 + i)(x + iy) = 2x + ix + 2iy - y so there are two real parts 2x and -y and I don't know which one they want" or "the x was the real part and 2x is the only x without an i so 2x is the real part". They aren't asking for depth; just comprehension. Which you have. $\endgroup$
    – fleablood
    Commented Jan 18, 2016 at 18:08

2 Answers 2


To address your update: what you saw on MathWorld is correct: if $z = a + b {\rm i}$, then $\bar z = a - b {\rm i}$, so $$\frac {z + \bar z} 2 = \frac {a + b {\rm i} + a - b {\rm i}} 2 = a = \operatorname{Re} z ,$$ so there is no contradiction. A similar calculation can be done for $\operatorname{Im} z$.

  • $\begingroup$ Omg....thank you.....I can't believe I didn't notice that. You rock. Don't ever change. $\endgroup$ Commented Jan 18, 2016 at 20:59

You're totally right!


$$\Re\left((2+i)(x+iy)\right)=\Re\left(2x+2iy+ix+iiy\right)=$$ $$\Re\left(2x+2iy+ix+i^2y\right)=\Re\left(2x+2iy+ix-y\right)=$$ $$\Re\left(2x-y+2iy+ix\right)=\Re\left((2x-y)+(2y+x)i\right)=2x-y$$

And, so:


  • $\begingroup$ $(2y+x)i$ would be correct, no? For the imaginary part. You seem to have left out the $i$. $\endgroup$ Commented Jan 18, 2016 at 21:19
  • $\begingroup$ No, the $\Im(z)$ means the part with the $i$ part so you've to leave the $i$ out $\endgroup$ Commented Jan 18, 2016 at 21:27
  • $\begingroup$ Oh ok, thanks!! $\endgroup$ Commented Jan 18, 2016 at 23:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .