There is a lot of confusion in the question, confusion that could be reduced by looking at a textbook on the paper, or even the original 1977 paper by Dempster, Laird and Rubin.
Here is an excerpt of our book, Introducing Monte Carlo Methods with R, followed by my answer:
Assume that we observe $X_1, \ldots, X_n$, jointly distributed from $g({\mathbf x}|\theta)$ that satisfies
$$
g({\mathbf x}|\theta)=\int_{\cal Z} f({\mathbf x}, {\mathbf z}|\theta)\, \text{d}{\mathbf z},
$$
and that we want to compute $\hat{\theta} = \arg\max L(\theta|{\mathbf x})= \arg\max g({\mathbf x}|\theta)$.
Since the augmented data is ${\mathbf z}$, where $({\mathbf X}, {\mathbf Z}) \sim f({\mathbf x},{\mathbf z}| \theta)$
the conditional distribution of the missing data ${\mathbf Z}$ given the observed data ${\mathbf x}$ is
$$
k({\mathbf z}| \theta, {\mathbf x}) = f({\mathbf x}, {\mathbf z}|\theta)\big/g({\mathbf x}|\theta)\,.
$$
Taking the logarithm of this expression
leads to the following relationship between the complete-data likelihood $L^c(\theta|{\mathbf x},
{\mathbf z})$ and the observed-data likelihood $L(\theta|{\mathbf x})$. For any value $\theta_0$,
$$
\log L(\theta|{\mathbf x})= \mathbb{E}_{\theta_0}[\log L^c(\theta|{\mathbf x},{\mathbf Z})]
-\mathbb{E}_{\theta_0}[\log k({\mathbf Z}| \theta, {\mathbf x})],\qquad(1)
$$where the expectation is with respect to $k({\mathbf z}| \theta_0, {\mathbf x})$. In the EM algorithm,
while we aim at maximizing $\log L(\theta|{\mathbf x})$, only the first term on the right side of
(1) will be considered.
Denoting$$
Q(\theta |\theta_0, {\mathbf x}) = \mathbb{E}_{\theta_0}
[\log L^c(\theta|{\mathbf x},{\mathbf Z})],
$$
the EM algorithm indeed proceeds iteratively by maximizing
$Q(\theta |\theta_0, {\mathbf x})$ at each iteration and, if $\hat{\theta}_{(1)}$
is the value of $\theta$ maximizing $Q(\theta |\theta_0, {\mathbf x})$,
by replacing $\theta_0$ by the updated value $\hat{\theta}_{(1)}$. In this manner, a sequence of estimators
$\{\hat{\theta}_{(j)}\}_j$ is obtained, where $\hat{\theta}_{(j)}$ is defined as the value of
$\theta$ maximizing $Q(\theta |\hat{\theta}_{(j-1)}, {\mathbf x})$; that is,$$
Q(\hat{\theta}_{(j)} |\hat{\theta}_{(j-1)}, {\mathbf x})
= \max_{\theta}\,Q(\theta |\hat{\theta}_{(j-1)},
{\mathbf x}).$$This iterative scheme thus contains both an expectation step
and a maximization step, giving the algorithm its name.
EM Algorithm
Pick a starting value $\hat{\theta}_{(0)}$
Repeat
Compute the E-step
$$
Q(\theta|\hat{\theta}_{(m)}, {\mathbf x})
=\mathbb{E}_{\hat{\theta}_{(m)}} [\log L^c(\theta|{\mathbf x}, {\mathbf Z})]\,,
$$
where the expectation is with respect to $k({\mathbf z}|\hat{\theta}_{(m)},{\mathbf x})$ and set $m=0$.
Maximize $Q(\theta|\hat{\theta}_{(m)}, {\mathbf x})$ in
$\theta$ and take the M-step
$$
\hat\theta_{(m+1)}=\arg\max_\theta \; Q(\theta|\hat{\theta}_{(m)}, {\mathbf x})
$$
and set $m=m+1$
until a fixed point is reached; i.e., $\hat\theta_{(m+1)}=\hat{\theta}_{(m)}$.
For the normal mixture, using the missing data structure exhibited in previously leads to an objective function
equal to
$$
Q(\theta^\prime|\theta,\mathbf{x}) = -\frac{1}{2}\,\sum_{i=1}^n
\mathbb{E}_\theta\left[\left. Z_i (x_i-\mu_1)^2 + (1-Z_i) (x_i-\mu_2)^2 \right| \mathbf{x} \right].
$$
Solving the M-step then provides the closed-form expressions
$$
\mu_1^\prime = \mathbb{E}_\theta\left[ \sum_{i=1}^n Z_i x_i |\mathbf{x} \right]
\bigg/ \mathbb{E}_\theta\left[ \sum_{i=1}^n Z_i| \mathbf{x} \right]
$$
and
$$
\mu_1^\prime = \mathbb{E}_\theta\left[ \sum_{i=1}^n (1-Z_i) x_i |\mathbf{x} \right]
\bigg/ \mathbb{E}_\theta\left[ \sum_{i=1}^n (1-Z_i)| \mathbf{x} \right].
$$
Since
$$
\mathbb{E}_\theta\left[Z_i|\mathbf{x} \right]=\frac{\varphi(x_i-\mu_1)}{ \varphi(x_i-\mu_1)+3\varphi(x_i-\mu_2)}\,,
$$
the EM algorithm can easily be implemented in this setting.
Whatever the mixture involved, the latent variables $Z_i$ are Multinomial $\mathcal{M}_M(1;\pi_1,\ldots,\pi_M)$ which means only one component of the vector $Z_i$ is equal to one and all of the $M-1$ others are zero. (Note the difference with the question in the notations: the original notation $\mathcal{M}(M;\pi_1,\ldots,\pi_M)$ fails to indicate how many draws are taken, that is, what is the sum of the components of $Z_i$.).When $k=2$ as in the above excerpt, $Z_i$ is an integer in $\{0,1\}$. There may be a confusion between a Multinomial distribution and the property of a distribution (like some mixtures) to be multimodal. The $Z_i$ do not have a multimodal distribution, taking only two values, even conditional on the $X_i$'s, while the $X_i$'s may, at least unconditionally.