The numbers z_1, z_2, z_3 are such that…

The numbers $z_1, z_2, z_3$ are such that: $$ \arg(z_1)=\arg(z_3)=\frac\pi6; \quad \arg(z_2)=-\frac\pi3; \quad |z_1|=1; \quad |z_3|=\frac{\sqrt2}{2}. $$ Determine $z_2$ such that $\text{Im}(z_1+z_2+z_3)=0$.

Explanation of Method

$z_1$ and $z_3$ are fully known (modulus and argument both given), so compute them directly in Cartesian form. $z_2$ has a known direction but unknown modulus $r=|z_2|$, so write $z_2=r\,e^{-i\pi/3}$ and add all three. The condition $\text{Im}(z_1+z_2+z_3)=0$ becomes one linear equation in the single unknown $r$.

Step 1 · Compute $z_1$

$$ z_1 = 1\cdot e^{i\pi/6} = \cos\frac\pi6+i\sin\frac\pi6 = \frac{\sqrt3}{2}+i\frac12 $$

Step 2 · Compute $z_3$

$$ z_3 = \frac{\sqrt2}{2}e^{i\pi/6} = \frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}+i\frac12\right) = \frac{\sqrt6}{4}+i\frac{\sqrt2}{4} $$

Step 3 · Write $z_2$ with unknown modulus $r$

Using $\cos\left(-\frac\pi3\right)=\frac12$, $\sin\left(-\frac\pi3\right)=-\frac{\sqrt3}{2}$:

$$ z_2 = r\,e^{-i\pi/3} = \frac{r}{2}-i\frac{r\sqrt3}{2} $$

Step 4 · Find the imaginary part of the sum

$$ \text{Im}(z_1+z_2+z_3) = \frac12 - \frac{r\sqrt3}{2}+\frac{\sqrt2}{4} = \frac{2-2r\sqrt3+\sqrt2}{4} $$

Step 5 · Solve for $r$

Setting the imaginary part to zero:

$$ 2-2r\sqrt3+\sqrt2 = 0 \quad\Longrightarrow\quad r = \frac{2+\sqrt2}{2\sqrt3} $$

Step 6 · Rationalize

$$ r = \frac{2+\sqrt2}{2\sqrt3}\cdot\frac{\sqrt3}{\sqrt3} = \frac{2\sqrt3+\sqrt6}{6} $$

Step 7 · Write $z_2$ in Cartesian form

$$ z_2 = r\left(\frac12-i\frac{\sqrt3}{2}\right) = \frac{2\sqrt3+\sqrt6}{12} - i\,\frac{2+\sqrt2}{4} $$

Step 8 · Verify

$$ \text{Im}(z_1)+\text{Im}(z_2)+\text{Im}(z_3) = \frac12 - \frac{2+\sqrt2}{4} + \frac{\sqrt2}{4} = \frac{2-2-\sqrt2+\sqrt2}{4} = 0 \quad\checkmark $$

Step 9 · Numeric check

$$ z_1\approx0.866+0.5i, \qquad z_3\approx0.612+0.354i, \qquad r\approx0.986, \qquad z_2\approx0.493-0.854i $$

$$ z_1+z_2+z_3 \approx 1.971 + 0.000i \quad\checkmark $$

$z_1$ (angle $30°$, radius $1$) and $z_3$ (angle $30°$, radius $\frac{\sqrt2}{2}$) both point up-right; $z_2$ (angle $-60°$, radius $r\approx0.986$, not shown to scale) is tuned so all three imaginary parts cancel, leaving a purely real sum.

Result

$$ z_2 = \frac{2+\sqrt2}{2\sqrt3}\,e^{-i\pi/3} = \frac{2\sqrt3+\sqrt6}{12} - i\,\frac{2+\sqrt2}{4} $$

with $|z_2| = \dfrac{2+\sqrt2}{2\sqrt3} = \dfrac{2\sqrt3+\sqrt6}{6} \approx 0.986$, chosen precisely so that $\text{Im}(z_1+z_2+z_3)=0$ (verified both symbolically and numerically).