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The question:

Calculate the inverse for the function from $\mathbb{C}$ to $\mathbb{C}$ given by $a + bi \mapsto 2a - b^3i$. Use the inverse function to show that the given function is surjective.


The part where I'm stuck is to find a function that undoes the action of the above function, to be more exact is that I don't know if I should treat $a$ and $b$ as two different variables or I should treat them as a whole. To show the function is surjective I can compose the function and its inverse to get the identity (which i'm not sure what to expect) then compose the inverse with the function and I should get the same identity and this prove that the function is bijective thus it is surjective.

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  • $\begingroup$ Hint: In $\;\mathbb{C}\;$ $\;a+b i=a'+b'i \;\text{with}\; a,b,a',b' \in \mathbb{R}\; \;\iff\; a=a' \;\text{and}\; b=b'$ so $a,b$ are independent variables.. $\endgroup$ – dxiv Oct 16 '16 at 17:56
  • $\begingroup$ Hint #2: the real function $a \mapsto 2 a$ is surjective and its inverse is $a' \mapsto \frac{1}{2} a'$. Now look at $b \mapsto -b^3$. $\endgroup$ – dxiv Oct 16 '16 at 18:59
  • $\begingroup$ I kinda got the first part, where I treat $a$ and $b$ separately. However the second part is the issue right now. From what I understood from your 2nd hint is that I find the inverse of $a$ and $b$ separately then prove the inverse of $a$ and $b$ are surjective and by proving these two statements it means that the function is surjective. Am I missing something or I'm on the right track? $\endgroup$ – Zed Oct 16 '16 at 19:10
  • $\begingroup$ Yes, that's the idea. $\endgroup$ – dxiv Oct 16 '16 at 19:12

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