Well, I am not getting any hint how to show $GL_n(\mathbb{C})$ is path connected. So far I have thought that let $A$ be any invertible complex matrix and $I$ be the idenity matrix, I was trying to show a path from $A$ to $I$ then define $f(t)=At+(1-t)I$ for $t\in[0,1]$ which is possible continous except where the $\operatorname{det}{f(t)}=0$ i.e. which has $n$ roots and I can choose a path in $\mathbb{C}\setminus\{t_1,\dots,t_n\}$ where $t_1,\dots,t_n$ are roots of $\operatorname{det}{f(t)}=0$, is my thinking was correct? Could anyone tell me the solution?
Since any matrix $A\in GL_n(\mathbb C)$ has only finitely many eigenvalues, and 0 isn’t one of them, there is a point $z\in S^1$ such that the line through the origin containing $z$ doesn’t intersect any of the eigenvalues of $A$. Now, consider the path $f(t)=At+z(1-t)I$. This has determinant 0 iff $z(t-1)$ is an eigenvalue of $At$, which happens iff $z(1-1/t)$ is an eigenvalue of $A$ (this doesn’t work when $t=0$, but then it is clear that the determinant is non-zero). By construction, it isn’t for any $t\in[0,1]$ so this defines a path form $A$ to $zI$. now there is a path not passing through 0 from $z$ to 1, and this gives rise to a path from $zI$ to $I$, and so concatenating the two paths, we get a path from $A$ to $I$, showing that $GL_n(\mathbb C)$ is path connected.
Use $\Gamma(t) = e^{t \log A + (1-t) \log B}$. This is well defined since $A,B$ are invertible. $\Gamma(t)$ is clearly invertible for all $t$, $\Gamma(0) = B$, $\Gamma(1) = A$.
Following idea shows connectedness of $Gl(n,\mathbb{C})$ ..
As $A\in Gl(n,\mathbb{C})$ you do have an upper triangular matrix which is similar to $A$.
See subgroup of Invertible, Upper triangular matrices as
$$\underbrace{\mathbb{C}^*\times\mathbb{C}^*\times \cdots\times\mathbb{C}^*}_{n- times}\times \underbrace{\mathbb{C}\times\mathbb{C}\times \mathbb{C}\times \cdots \times\mathbb{C}}_{\dfrac{n(n-1)}{2}times}$$
As $\mathbb{C}^*\times\mathbb{C}^*\times \cdots\times\mathbb{C}^*$ is connected and $\mathbb{C}$ is connected so is the above product.
See that conjugation is continuous so preserves connectedness.
By which i mean that $\{BUB^{-1} : U – \text{Upper Triangular}\}$ is connected.
$Gl(n,\mathbb{C})$ being union of connected sets is also connected.
See that all these conjugates have Identity matrix in common.
I think the answer by Praphulla Koushik has the (to me) most elementary approach, but it is awkwardly formulated, so I will try to restate it more simply. It suffices to check two statements:
The group $B$ of invertible upper triangular matrices is path-connected.
The union of conjugates of $B$ (each of which contains $I$) equals $GL(n,\Bbb C)$.
Then every $A\in GL(n,\Bbb C)$ can be joined to $I$ by a path that stays within a conjugate of $B$ that contains$~A$.
For point 1. it suffices to observe that $B$ is homeomorphic to the Cartesian product of the group $T$ of invertible diagonal matrices and the set $U$ of strictly upper triangular matrices; the former is a power of $\Bbb C^*$, which is path-connected, and the latter to a power of$~\Bbb C$.
Point 2. follows from the well known fact that any complex matrix is trigonalisable. This is equivalent to saying that every linear operator$~T$ on $\Bbb C^n$ admits a complete flag of $T$-stable subspaces. To find one, one chooses a left-eigenvector (a linear form on$~\Bbb C^n$) of$~T$, whose kernel is a $T$-stable hyperplane$~H$, and applies induction on the dimension to the restriction of $T$ to$~H$ to find the remaining $T$-stable subspaces for the flag.
By singular value decomposition and the fact that every unitary matrix is unitarily diagonalisable, we can write every invertible matrix $A$ as a product of the form $(US_1U^\ast)S_2(VS_3V^\ast)$, where $U,V$ are unitary matrices and $S_1,S_2,S_3$ are invertible diagonal matrices. So, it suffices to show that each such diagonal matrix is path-connected to $I$, and this boils down to the scalar case, which is trivial.