Let a, b be two real numbers such that ab

Question

(D) Given $\left|\frac{1+ai}{b+i}\right| = 1$,we have $|1+ai| = |b+i|$.Squaring both sides,$1+a^2 = b^2+1$,which implies $a^2 = b^2$,so $a = \pm b$.Si

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) Given $\left|\frac{1+ai}{b+i}ight| = 1$,we have $|1+ai| = |b+i|$.Squaring both sides,$1+a^2 = b^2+1$,which implies $a^2 = b^2$,so $a = \pm b$.Since $ab Given $a+ib$ lies on $|z-1| = |2z|$,we have $|(a-1)+ib| = 2|a+ib|$.Squaring both sides,$(a-1)^2 + b^2 = 4(a^2 + b^2)$.Substituting $b^2 = a^2$,we get $(a-1)^2 + a^2 = 4(2a^2) = 8a^2$.$a^2 - 2a + 1 + a^2 = 8a^2 \Rightarrow 6a^2 + 2a - 1 = 0$.Solving for $a$,$a = \frac{-2 \pm \sqrt{4 - 4(6)(-1)}}{12} = \frac{-2 \pm \sqrt{28}}{12} = \frac{-1 \pm \sqrt{7}}{6}$.If $a = \frac{\sqrt{7}-1}{6} \approx 0.27$,then $[a] = 0$ and $b = -a = \frac{1-\sqrt{7}}{6}$.Then $\frac{1+[a]}{4b} = \frac{1}{4(\frac{1-\sqrt{7}}{6})} = \frac{6}{4(1-\sqrt{7})} = \frac{3}{2(1-\sqrt{7})} 	imes \frac{1+\sqrt{7}}{1+\sqrt{7}} = \frac{3(1+\sqrt{7})}{2(1-7)} = \frac{3(1+\sqrt{7})}{-12} = -\frac{1+\sqrt{7}}{4}$.If $a = \frac{-1-\sqrt{7}}{6} \approx -0.607$,then $[a] = -1$ and $b = -a = \frac{1+\sqrt{7}}{6}$.Then $\frac{1+[a]}{4b} = \frac{1+(-1)}{4b} = 0$.

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

Similar Questions

If $i = \sqrt{-1}$, then $\frac{e^{xi} + e^{-xi}}{2} = $

If $z=x+iy$ is a complex number such that $\bar{z}^{\frac{1}{3}}=a+ib$,then the value of $\frac{1}{a^2+b^2}\left(\frac{x}{a}+\frac{y}{b}\right)$ is equal to

The value of $\left(\frac{-1+i \sqrt{3}}{1-i}\right)^{30}$ is

If $a_k = \cos \alpha_k + i \sin \alpha_k$ for $k = 1, 2, 3$ and $a_1, a_2, a_3$ are the roots of the equation $x^3 + bx + c = 0$,then the real part of $b$ is:

If $z = x - iy$ and $z^{1/3} = p + iq$ $(x, y, p, q \in R)$, then $\frac{(\frac{x}{p} + \frac{y}{q})}{(p^2 + q^2)}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module