Solve the equation e(fracπ4)z + e(fracπ6)z + i = 0, where…

Solve the equation $$ e\!\left(\frac\pi4\right)z + e\!\left(\frac\pi6\right)z + i = 0, \qquad \text{where } e(\alpha)=e^{i\alpha}. $$

Step 1 · Convert to Cartesian form

$$ e^{i\pi/4} = \frac{\sqrt2}{2}+i\frac{\sqrt2}{2}, \qquad e^{i\pi/6} = \frac{\sqrt3}{2}+i\frac12 $$

Step 2 · Factor out $z$

$$ \left(e^{i\pi/4}+e^{i\pi/6}\right)z = -i $$

Step 3 · Add the two exponentials

$$ e^{i\pi/4}+e^{i\pi/6} = \frac{\sqrt2+\sqrt3}{2} + i\,\frac{\sqrt2+1}{2} = \frac12\left[(\sqrt2+\sqrt3)+i(\sqrt2+1)\right] $$

Step 4 · Solve for $z$

$$ z = \frac{-i}{\frac12\left[(\sqrt2+\sqrt3)+i(\sqrt2+1)\right]} = \frac{-2i}{(\sqrt2+\sqrt3)+i(\sqrt2+1)} $$

Step 5 · Rationalize the denominator

Multiply by the conjugate $(\sqrt2+\sqrt3)-i(\sqrt2+1)$. The denominator becomes:

$$ (\sqrt2+\sqrt3)^2+(\sqrt2+1)^2 = (5+2\sqrt6)+(3+2\sqrt2) = 8+2\sqrt6+2\sqrt2 = 2(4+\sqrt6+\sqrt2) $$

The numerator, using $i^2=-1$:

$$ -2i\left[(\sqrt2+\sqrt3)-i(\sqrt2+1)\right] = -2i(\sqrt2+\sqrt3) + 2i^2(\sqrt2+1) = -2(\sqrt2+1) - 2i(\sqrt2+\sqrt3) $$

Step 6 · Simplify

$$ z = \frac{-2(\sqrt2+1)-2i(\sqrt2+\sqrt3)}{2(4+\sqrt6+\sqrt2)} = -\frac{(\sqrt2+1)+i(\sqrt2+\sqrt3)}{4+\sqrt6+\sqrt2} $$

$$ \boxed{z = -\frac{\sqrt2+1}{4+\sqrt6+\sqrt2} - i\,\frac{\sqrt2+\sqrt3}{4+\sqrt6+\sqrt2}} $$

Step 7 · Numeric check

Using $\sqrt2\approx1.414$, $\sqrt3\approx1.732$, $\sqrt6\approx2.449$: denominator $\approx7.863$, giving:

$$ z \approx -0.307 - 0.400i $$

Substituting back: $e^{i\pi/4}z+e^{i\pi/6}z+i \approx 0$ (verified numerically to machine precision).