Let a,b be two real numbers such that ab

Question

Let $a,b$ be two real numbers such that $ab<0$. If the complex number $\dfrac{1+ai}{b+i}$ is of unit modulus and $a+ib$ lies on the circle $|z-1|=|2z|$, then a possible value of $\dfrac{1+[a]}{4b}$, where $[t]$ is the greatest integer function, is:

Accepted Answer

From $\left|\dfrac{1+ai}{b+i}\right|=1$: $|1+ia|=|b+i| \implies a^2+1=b^2+1 \implies a=\pm b$. Since $ab<0$, we need $b=-a$.

From $|a+ib-1|=2|a+ib|$: $(a-1)^2+b^2=4(a^2+b^2)$. With $b^2=a^2$:
$$(a-1)^2=4(2a^2) \implies 1-2a=6a^2 \implies 6a^2+2a-1=0.$$

$$a=\frac{-2\pm\sqrt{28}}{12}=\frac{-1\pm\sqrt7}6.$$

Case $a=\dfrac{\sqrt7-1}6$: $[a]=0$, and $b=\dfrac{1-\sqrt7}6$, giving $\dfrac{1+[a]}{4b}=\dfrac{1}{4\left(\frac{1-\sqrt7}6\right)}=-\dfrac{1+\sqrt7}4$ (not among the neat options).

Case $a=\dfrac{-1-\sqrt7}6$: $[a]=-1$, and $b=\dfrac{1+\sqrt7}6$, giving $\dfrac{1+[a]}{4b}=\dfrac{0}{4b}=0$.

Solutions · 1