i am asking too simple question, sorry for that. what is the difference between these two imaginär numbers?

$\operatorname{Im}(| \sqrt2+3i|^2)$ vs. $\operatorname{Im}((\sqrt2+3i)^2)$

$| \sqrt2+3i|^2$ means $ \sqrt2^2+3i^2$ right?

  • 1
    $\begingroup$ $\vert \sqrt{2}+3i\vert^2=\sqrt{2}^2+3^2$, no $i$. $\endgroup$
    – icurays1
    Commented Dec 1, 2012 at 16:54
  • $\begingroup$ oh yeah, thats right. $\endgroup$
    – doniyor
    Commented Dec 1, 2012 at 16:56

1 Answer 1


Short answer: the first one is zero since $\vert \sqrt{2}+3i\vert^2$ is a real number (it's the modulus of your complex number).

Long answer: compute them and see! Recall that $\vert x+iy\vert=\sqrt{x^2+y^2}$:

$$ \vert \sqrt 2+3i\vert^2=\sqrt{2}^2+3^2=11\\ (\sqrt{2}+3i)^2=\sqrt{2}^2+(3i)^2+2(\sqrt{2}\cdot 3i)=-7+6\sqrt{2}i $$

So the imaginary part of the second number is $6\sqrt{2}$.

  • $\begingroup$ yeah, you are right. thanks. $\endgroup$
    – doniyor
    Commented Dec 1, 2012 at 16:53

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