Part A: Given complex numbers z_1 = cos(t)/(2) + isin(t)/(2)…
Part A: Given complex numbers $z_1 = \cos\dfrac{t}{2} + i\sin\dfrac{t}{2}$ and $z_2 = \dfrac{\sqrt{2}}{2}\sin t + i$, determine the real parameter $t$ from the condition $z_1 = z_2$.
Part B: Given complex numbers $z_1 = 1 + 2ni$ and $z_2 = 3n - 4i$, determine the real parameter $n$ such that $\text{Im}(z_1+z_2) = 0$.
Explanation of Method
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. In Part A this gives two equations in $t$ to solve simultaneously; in Part B the sum $z_1+z_2$ is first written in standard form $x+iy$, and the condition $\text{Im}(z_1+z_2)=0$ becomes a single linear equation in $n$.
Part A · Finding $t$ from $z_1 = z_2$
Step 1 · Write the condition
$$ z_1 = \cos\frac{t}{2} + i\sin\frac{t}{2}, \qquad z_2 = \frac{\sqrt{2}}{2}\sin t + i $$
For $z_1 = z_2$, both the real and imaginary parts must match.
Step 2 · Equate real parts
$$ \text{Re}(z_1) = \text{Re}(z_2) \quad\Longrightarrow\quad \cos\frac{t}{2} = \frac{\sqrt{2}}{2}\sin t \qquad \text{(1)} $$
Step 3 · Equate imaginary parts
$$ \text{Im}(z_1) = \text{Im}(z_2) \quad\Longrightarrow\quad \sin\frac{t}{2} = 1 \qquad \text{(2)} $$
Step 4 · Solve equation (2)
$$ \sin\frac{t}{2} = 1 \quad\Longrightarrow\quad \frac{t}{2} = \frac{\pi}{2} + 2\pi k, \quad k \in \mathbb{Z} \quad\Longrightarrow\quad t = \pi + 4\pi k, \quad k \in \mathbb{Z} $$
The principal value is $t = \pi$.
Step 5 · Check against equation (1)
Substituting $t = \pi$:
$$ \cos\frac{\pi}{2} = \frac{\sqrt{2}}{2}\sin\pi \quad\Longrightarrow\quad 0 = \frac{\sqrt{2}}{2}\cdot 0 \quad\Longrightarrow\quad 0 = 0 \ \checkmark $$
Verification: at $t = \pi$: $z_1 = \cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2} = i$, and $z_2 = \dfrac{\sqrt2}{2}\sin\pi + i = i$. Indeed $z_1 = z_2 = i$. ✓
$$ \boxed{t = \pi + 4\pi k,\ k \in \mathbb{Z} \qquad \text{(principal value: } t=\pi\text{)}} $$
Part B · Finding $n$ such that $\text{Im}(z_1+z_2) = 0$
Step 1 · Compute $z_1 + z_2$
$$ z_1 + z_2 = (1 + 2ni) + (3n - 4i) = (1 + 3n) + i(2n - 4) $$
Step 2 · Extract the imaginary part
$$ \text{Im}(z_1+z_2) = 2n - 4 $$
Step 3 · Apply the condition
$$ 2n - 4 = 0 \quad\Longrightarrow\quad n = 2 $$
Verification: for $n=2$: $z_1 = 1+4i$, $z_2 = 6-4i$, so $z_1+z_2 = 7 + 0i = 7$, a purely real number. ✓
$$ \boxed{n = 2} $$
Summary of Methods
Part A: equate real and imaginary parts separately; solving $\sin\dfrac{t}{2}=1$ gives $\dfrac{t}{2}=\dfrac{\pi}{2}+2\pi k$; then verify the solution satisfies both equations.
Part B: compute the sum $z_1+z_2$, extract its imaginary part, and set that equal to zero to solve for $n$.