Problem 230

To solve for $\lambda$, we equate the real and imaginary parts of $z_1$ and $z_2$ according to the property that two complex numbers are equal if and only if their corresponding real and imaginary parts are equal.

From the real parts, we have:
$$m = 2\cos \theta$$
From the imaginary parts, we have:
$$4 - m^2 = \lambda + 3\sin \theta$$
Substitute $m = 2\cos \theta$ into the second equation:
$$\lambda = 4 - 4\cos^2 \theta - 3\sin \theta$$
Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we can rewrite the equation as:
$$\lambda = 4 - 4(1 - \sin^2 \theta) - 3\sin \theta$$
$$= 4\sin^2 \theta - 3\sin \theta$$
To simplify further, complete the square for $\sin \theta$:
$$\lambda = 4\left( \sin \theta - \frac{3}{8} \right)^2 - \frac{9}{16}$$

Since the sine function is bounded between -1 and 1, i.e., $-1 \leq \sin \theta \leq 1$, the minimum and maximum of the quadratic expression in the parentheses occur either at the vertex or at one of the endpoints of the interval.

The vertex of $\left( \sin \theta - \frac{3}{8} \right)^2$ is at $\sin \theta = \frac{3}{8}$ which gives the minimum value for $\lambda$:
$$\lambda_{\min} = - \frac{9}{16}$$

The maximum value for $\lambda$ occurs at the endpoint where $\sin \theta = -1$:
$$\lambda_{\max} = 4\left( -1 - \frac{3}{8} \right)^2 - \frac{9}{16} = 7$$

Thus, the range of possible values for $\lambda$ is:
$$\boxed{\left[- \frac{9}{16}, 7\right]}$$