If k is a real positive number, determine the complex…
If $k$ is a real positive number, determine the complex numbers $z_1$ and $z_2$ from the conditions: $$ \frac{z_1}{\bar z_2} = i\sqrt2, \qquad z_1\bar z_2 = k. $$
Explanation of Method
Treat $\bar z_2$ as the working unknown. The ratio condition expresses $z_1$ directly in terms of $\bar z_2$; substituting into the product condition gives a single quadratic equation for $\bar z_2$, solved by converting its right-hand side to polar form and taking a square root. Once $\bar z_2$ is known, both $z_2$ (by conjugating) and $z_1$ (by the ratio relation) follow.
Step 1 · Express $z_1$ in terms of $\bar z_2$
$$ \frac{z_1}{\bar z_2} = i\sqrt2 \quad\Longrightarrow\quad z_1 = i\sqrt2\,\bar z_2 $$
Step 2 · Substitute into the product condition
$$ z_1\bar z_2 = k \quad\Longrightarrow\quad i\sqrt2\,(\bar z_2)^2 = k \quad\Longrightarrow\quad (\bar z_2)^2 = \frac{k}{i\sqrt2} $$
Step 3 · Simplify the right-hand side
Multiplying numerator and denominator by $-i$ (since $\frac1i=-i$):
$$ \frac{k}{i\sqrt2} = \frac{-ik}{\sqrt2} = -i\,\frac{k\sqrt2}{2} $$
so $(\bar z_2)^2 = -i\,\dfrac{k\sqrt2}{2}$, a purely imaginary number with negative imaginary part.
Step 4 · Take the square root in polar form
Write $w = -i\dfrac{k\sqrt2}{2}$: its modulus is $|w|=\dfrac{k\sqrt2}{2}$ and its argument is $-\dfrac\pi2$ (negative imaginary axis), so $w = \dfrac{k\sqrt2}{2}\,e^{-i\pi/2}$. Then:
$$ \bar z_2 = \pm\sqrt{\frac{k\sqrt2}{2}}\;e^{-i\pi/4} $$
Simplifying the modulus: $\sqrt{\dfrac{k\sqrt2}{2}} = \sqrt{k}\cdot\sqrt{2^{-1/2}} = \sqrt{k}\cdot2^{-1/4} = \dfrac{\sqrt k}{\sqrt[4]{2}}$.
Step 5 · Write $\bar z_2$ and $z_2$ explicitly
Using $e^{-i\pi/4}=\dfrac{\sqrt2}{2}-i\dfrac{\sqrt2}{2} = \dfrac{\sqrt2}{2}(1-i)$:
$$ \bar z_2 = \pm\frac{\sqrt k}{\sqrt[4]2}\cdot\frac{\sqrt2}{2}(1-i) = \pm\frac{\sqrt k}{2^{3/4}}(1-i) $$
Conjugating gives $z_2$:
$$ z_2 = \pm\frac{\sqrt k}{2^{3/4}}(1+i) $$
Step 6 · Find $z_1$
Using $z_1=i\sqrt2\,\bar z_2$ and $i(1-i)=i-i^2=1+i$:
$$ z_1 = i\sqrt2\cdot\left[\pm\frac{\sqrt k}{2^{3/4}}(1-i)\right] = \pm\frac{\sqrt{2k}}{2^{3/4}}(1+i) = \pm\frac{\sqrt k}{2^{1/4}}(1+i) $$
Step 7 · Summary of the two solutions
$$ z_1 = \pm\frac{\sqrt k}{\sqrt[4]{2}}(1+i), \qquad z_2 = \pm\frac{\sqrt k}{2^{3/4}}(1+i) \qquad \text{(same sign in both)} $$
Step 8 · Verify (taking the $+$ sign)
$$ \bar z_2 = \frac{\sqrt k}{2^{3/4}}(1-i), \qquad \frac{z_1}{\bar z_2} = \frac{2^{3/4}}{2^{1/4}}\cdot\frac{1+i}{1-i} = \sqrt2\cdot\frac{1+i}{1-i} $$
Rationalizing: $\dfrac{1+i}{1-i} = \dfrac{(1+i)^2}{(1-i)(1+i)} = \dfrac{2i}{2} = i$, so $\dfrac{z_1}{\bar z_2} = \sqrt2\cdot i = i\sqrt2$. ✓
$$ z_1\bar z_2 = \frac{k}{2^{1/4+3/4}}(1+i)(1-i) = \frac{k}{2}(1-i^2) = \frac{k}{2}\cdot2 = k \quad\checkmark $$
Both $z_1$ and $z_2$ point in the same direction (along $1+i$, angle $45°$); only their lengths differ, by a factor of $\sqrt2$.
Result
Two solutions (matching signs), both verified against $\dfrac{z_1}{\bar z_2}=i\sqrt2$ and $z_1\bar z_2=k$. Note $|z_1|=\sqrt2\,|z_2|$, and both numbers point in the direction of $1+i$.