Prove that $\mathbb{C}^*$ is isomorphic to a subgroup of $GL_2(\mathbb{R})$ 
Prove that $\mathbb{C}^*$ is isomorphic to the subgroup of $GL_2(\mathbb{R})$ consisting of matrices
  of the form 
  $$\begin{bmatrix}
a & b\\ 
-b &  a
\end{bmatrix}$$

I showed that $\mathbb{C}^*$ and $GL_2(\mathbb{R})$ are homomorphic by proving that $\Phi(z$ o  $w)$ = $\Phi (z)$ . $\Phi (w)$ where $z,w$ are elements of  $\mathbb{C}^*$.
One-to one? I have shown that for any two complex $z$ and $w$ when $$\Phi (z) =\Phi (w) => z=w$$
Onto? 
Let
$$\begin{bmatrix}
m & n\\ 
p &  q
\end{bmatrix} \in GL_2(\mathbb{R})$$
and let $$\Phi (a+bi) = \begin{bmatrix} m & n \\ p & q\end {bmatrix}.$$ 
Then $$\begin{bmatrix}
a & b\\ 
-b &  a
\end{bmatrix} = \begin{bmatrix} m & n \\ p & q\end {bmatrix}$$ 
BUT then what can be said about that to prove that the relation is onto? i am a bit stuck here.
 A: Your map $\Phi$ isn't to $GL_2(\Bbb R)$ itself, but rather the subgroup  $G = \left\{\pmatrix{a & b \\ -b & a}: a, b \in \Bbb R\right\}$. The instructions aren't incredibly explicit, but it is very important here that $\Phi$ has codomain $G$, rather than all of $GL_2(\Bbb R)$.
When you pick a matrix $M =\pmatrix{m&n\\p&q}$, you're not necessarily picking something in $G$, the codomain of $\Phi$. It may very well be the case that there's no $z \in \Bbb C^*$ such that $\Phi(z) = M$; this happens if and only if $q = m$ and $p = -n$. So if you pick the $M$ above as your generic matrix and try to find some $z \in \Bbb C^*$ so that $\Phi(z) = M$, you had better not be able to show that $\Phi$ is onto $GL_2(\Bbb R)$ -- it's not!
Instead, to show that $\Phi$ is surjective ("onto"), pick something in the codomain of $\Phi$, something in $G$.
So pick a generic matrix $M = \pmatrix{m&n\\-n&m} \in G$, and find the corresponding $z \in \Bbb C^*$ with $\Phi(z) = M$.

In actuality, defining $\Phi$ as above but making its codomain all of $GL_2(\Bbb R)$ instead of just $G$ isn't a huge problem. It's a general fact that for any group homomorphism $\varphi: H \to H'$, if $\varphi$ is injective, then $H$ is isomorphic to the image $\operatorname{im}(\varphi) = \{\varphi(h) : h \in H\}$ (this is a consequence of the first isomorphism theorem).
Since you've shown that $\Phi$ is an injective homomorphism from $\Bbb C^*$, then $\Bbb C^*$ is necessarily isomorphic to $\operatorname{im}(\Phi)$, which is exactly the $G$ defined in the problem statement. It's just that, unless the codomain of $\Phi$ is exactly $\operatorname{im}(\Phi)$, you won't be able to show that $\Phi$ is surjective.
A: Note that GL$(2,R)$ is non-abelian and so any homomorphism from an abelian group cannot be onto. It would be interesting to describe the proper subgroup that is the image of the map you have described.Regarding complex numbers as two-dimensional real vector space, and multiplication with non-zero complex numbers being an invertible linear transformations, what you hvve done is writing down the matrices of the corresponding linear transformation.
