Determine the complex numbers z_1 and z_2 if: z_1+z_2 = 1-i…
Determine the complex numbers $z_1$ and $z_2$ if: $$ z_1+z_2 = 1-i, \qquad \arg(z_1)=\frac{\pi}{6}, \qquad \arg(z_2)=-\frac{\pi}{3}. $$
Explanation of Method
Since the arguments are fixed, each $z_k$ is an unknown positive real modulus $r_k$ times a known unit vector $e^{i\theta_k}$. Writing both in Cartesian form and adding turns the vector equation $z_1+z_2=1-i$ into two real linear equations in $r_1,r_2$ — solve that $2\times2$ system, then substitute back to get $z_1,z_2$ explicitly.
Step 1 · Write $z_1,z_2$ with unknown moduli
$$ z_1 = r_1e^{i\pi/6} = r_1\left(\cos\frac\pi6+i\sin\frac\pi6\right), \qquad z_2 = r_2e^{-i\pi/3} = r_2\left(\cos\frac\pi3-i\sin\frac\pi3\right) $$
with $r_1=|z_1|>0$, $r_2=|z_2|>0$ to be determined.
Step 2 · Plug in the trig values
$$ \cos\frac\pi6=\frac{\sqrt3}2,\ \ \sin\frac\pi6=\frac12, \qquad \cos\frac\pi3=\frac12,\ \ \sin\frac\pi3=\frac{\sqrt3}2 $$
$$ z_1 = \frac{r_1\sqrt3}{2}+i\frac{r_1}{2}, \qquad z_2 = \frac{r_2}{2}-i\frac{r_2\sqrt3}{2} $$
Step 3 · Add and match to $1-i$
$$ z_1+z_2 = \frac12(r_1\sqrt3+r_2) + \frac{i}{2}(r_1-r_2\sqrt3) = 1-i $$
gives the real system:
$$ r_1\sqrt3+r_2 = 2 \qquad (1) \qquad\qquad r_1-r_2\sqrt3 = -2 \qquad (2) $$
Step 4 · Solve for $r_1,r_2$
From (2): $r_1 = r_2\sqrt3-2$. Substituting into (1):
$$ \sqrt3(r_2\sqrt3-2)+r_2 = 2 \quad\Longrightarrow\quad 3r_2-2\sqrt3+r_2=2 \quad\Longrightarrow\quad 4r_2 = 2+2\sqrt3 $$
$$ r_2 = \frac{1+\sqrt3}{2} $$
Then:
$$ r_1 = r_2\sqrt3-2 = \frac{(1+\sqrt3)\sqrt3}{2}-2 = \frac{\sqrt3+3}{2}-2 = \frac{\sqrt3-1}{2} $$
Both are positive, as required for moduli. ✓
Step 5 · Check against the system
$$ r_1\sqrt3+r_2 = \frac{(\sqrt3-1)\sqrt3}{2}+\frac{1+\sqrt3}{2} = \frac{3-\sqrt3+1+\sqrt3}{2}=\frac42=2 \quad\checkmark $$
$$ r_1-r_2\sqrt3 = \frac{\sqrt3-1}{2}-\frac{(1+\sqrt3)\sqrt3}{2} = \frac{\sqrt3-1-\sqrt3-3}{2}=\frac{-4}{2}=-2 \quad\checkmark $$
Step 6 · Write $z_1,z_2$ explicitly
$$ z_1 = \frac{\sqrt3-1}{2}\left(\frac{\sqrt3}{2}+i\frac12\right) = \frac{3-\sqrt3}{4}+i\,\frac{\sqrt3-1}{4} $$
$$ z_2 = \frac{1+\sqrt3}{2}\left(\frac12-i\frac{\sqrt3}{2}\right) = \frac{1+\sqrt3}{4}-i\,\frac{\sqrt3+3}{4} $$
Step 7 · Verify the sum
Real parts: $\dfrac{3-\sqrt3+1+\sqrt3}{4}=\dfrac44=1$. Imaginary parts: $\dfrac{\sqrt3-1-\sqrt3-3}{4}=\dfrac{-4}{4}=-1$.
$$ z_1+z_2 = 1-i \quad\checkmark $$
$z_1$ at angle $30°$, then $z_2$ (at angle $-60°$, drawn tip-to-tail from $z_1$) lands exactly on $1-i$.
Result
with $|z_1|=\dfrac{\sqrt3-1}{2}\approx0.366$ and $|z_2|=\dfrac{1+\sqrt3}{2}\approx1.366$, verified to satisfy $z_1+z_2=1-i$ and to have the specified arguments $\dfrac\pi6$ and $-\dfrac\pi3$.