Let $S=\{(z_1,z_2,\ldots, z_n)\in \mathbb{C}^n: |z_1|^2+|z_2|^2+\ldots+|z_n|^2=1\}$ let $K=\{(z_1,z_2,\ldots, z_n)\in S: |z_1|^2+|z_2|^2+\ldots+|z_{n-1}|^2=1\}$. Prove that $S\setminus K=M$ has same homotopy type as circle?
My attempt: We define the deformation retract of $M$ to $\{(z_1,z_2,\ldots, z_n)\in S:|z_n|^2=1\}\subset M$ by
$$f_t=\frac{(1-t)(z_1,z_2,\ldots,z_n)+ t(0,0,\ldots, 0,\frac{z_{n}}{|z_n|} )}{|(1-t)(z_1,z_2,\ldots,z_n)+ t(0,0,\ldots, 0,\frac{z_{n}}{|z_n|} )|}$$
Will this homotopy work?
Are there any other ways to prove the result?
Yes it works. What details do you still need to check? Maybe:
This straight line homotopy in the numerator is never zero when $z_n \neq 0$, so it does make sense to normalize it. This follows from the next point.
The homotopy stays in $S\setminus K$, i.e. the last coordinate of the straight line homotopy is never zero if $z_n \neq 0$ . To check this note $(1-t)z_n + tz_n/|z_n| = 0$ is equivalent to $1 = t(1-1/|z_n|)$ but $t(1-1/|z_n|) \leq (1-1/|z_n|) \leq 1$