1) the set of diagonal matrices is path connected: if $A=\sum a_j E_{jj}$, $B=\sum b_j E_{jj}$ we take the map $t\mapsto \sum (ta_j+(1-t)b_j) E_{jj}$, $t\in[0,1]$.
2) The set of unitaries is path connected. If $U,V$ are two unitaries, we can always write them as $U=e^{iA}$, $V=e^{iB}$ with $A,B$ hermitian. Then we can consider the map $t\mapsto e^{itA}e^{i(1-t)B}$, $t\in[0,1]$ which gives a path from $U$ to $V$ within the unitary group.
3) The set of invertible hermitian matrices with positive eigenvalues is path connected. If $A,B$ are like that, then $A=UD_AU^*$, $B=VD_BV^*$. By parts 1) and 2), there exist continuous $f,g:[0,1]\to M_n(\mathbb{C})$ with $f(0)=D_A$, $f(1)=D_B$, $g(0)=U$, $g(1)=V$. Then $t\mapsto g(t)f(t)g(t)^*$ is continuous and takes $A$ to $B$. Note that the way that $f$ is defined guarantees that $f(t)$ will have positive eigenvalues for all $t\in[0,1]$.
4) GL$_n(\mathbb{C})$ is connected: Given $A,B$ invertible, we can write them as $A=RU$, $B=SV$ with $R,S$ hermitian and positive, and $U,V$ unitaries. By 3) and 2) we can find continuous functions $f,g:[0,1]\to M_n(\mathbb{C})$ with $f(0)=R$, $f(1)=S$, $g(0)=U$, $g(1)=V$. Then the map $t\mapsto f(t)g(t)$ is a continuous path from $A$ to $B$ (note that $f(t)$ and $g(t)$ are invertible for every $t\in[0,1]$ and then so is their product).
It only remains to justify the polar decomposition $A=RU$. An easy way to see this is by using the singular value decomposition. We write $A=WDV$, with $W,V$ unitaries and $D$ diagonal with non-negative entries (positive if $A$ is invertible). Then we can write
$$
A=(WDW^*)WV=RU,
$$
where $R=UDU^*$ is hermitian with positive eigenvalues (because $D$ is), and $U=WV$ is a unitary.
Best Answer
$$\begin{aligned} (A+iB)(A^\star-iB^\star) &=AA^\star+iBA^\star-iAB^\star + BB^\star\\ &=AA^\star +i(BA^\star-AB^\star) + BB^\star\\ &=A^2+B^2+i(BA^\star-AB^\star) \end{aligned}$$ Also $$\begin{aligned} (A^\star-iB^\star)(A+iB) &=AA^\star-iB^\star A+iA^\star B + BB^\star\\ &=AA^\star-i(B^\star A-A^\star B) + BB^\star\\ &=A^2+B^2-i(B A^\star-A B^\star) \end{aligned}$$ Since $U = A+iB$ is unitary, we know that $U^\star$ is its inverse and so we know that $I=UU^\star=U^\star U$ and so $$(A+iB)(A^\star-iB^\star)=(A^\star-iB^\star)(A+iB).$$ Therefore we can combine these pieces of information with the above derivations (subtract both) to conclude that $ 2i (BA^\star-AB^\star) = 0. $ Then we find $$ (A+iB)(A^\star-iB^\star) = A^2 + B^2. $$ Combine with $I=UU^\star$ and you have the result.