Complex Numbers
Argument condition; modulus of complex number
MJMT_Full_Test_03
Grade 12
Question:
If $Z$ is a complex number such that $\arg\left(z(1+\bar{z})\right) + \arg\left(\dfrac{|z|^2}{z - |z|^2 i}\right) = 0$, then
$\arg \bar{Z} = -\dfrac{\pi}{2}$
$\arg z = \dfrac{\pi}{4}$
$|z| < 1$
$\ln\left(\dfrac{1}{|z|}\right) \in (-\infty,\infty)$
Step-by-Step Solution
Key Concept: Combine the two arguments: $\arg\left(z(1+\bar{z}) \cdot \frac{z\bar{z}}{z-z\bar{z}\cdot i}\right)=0$. Simplify to show $\frac{1+\bar{z}}{1-\bar{z}}$ is real and positive.
$\arg z + \arg(1+\bar{z})+\arg\bar{z}-\arg(1-\bar{z})=0 \Rightarrow \arg\frac{1+\bar{z}}{1-\bar{z}}=0$. Hence $\frac{1+\bar{z}}{1-\bar{z}}=k>0$, so $\bar{z}=\frac{k-1}{k+1}\in(-1,1) \Rightarrow |z|<1$.
Correct Answer: 3