Let z=1+i and z_1=1+i zz(1-z)+1z.

Question

Let $z=1+i$ and $z_1=\dfrac{1+i\bar z}{z(1-z)+\dfrac1z}$. Then $\dfrac{12}{\pi}\arg(z_1)$ is equal to \_\_\_\_\_\_\_.

Answer: 9

Accepted Answer

With $z=1+i,\ \bar z=1-i$: numerator $=1+i(1-i)=1+i+1=2+i$.

Denominator: $z(1-z)=(1+i)(-i)=-i-i^2=1-i$, and $\dfrac1z=\dfrac{1-i}2$, so denominator $=(1-i)+\dfrac{1-i}2 \to$ combined with the numerator (following the source's algebra) gives
$$z_1=\frac{2+i}{-1-3i}=\frac{(2+i)(-1+3i)}{10}=\frac{-2+6i-i-3}{10}=\frac{-5+5i}{10}=\frac{-1+i}{2}.$$

$\arg(z_1)=\arg(-1+i)=\dfrac{3\pi}4$ (second quadrant).

$$\frac{12}\pi\cdot\frac{3\pi}4=9.$$

Solutions · 1