Let a, b be two real numbers such that ab < 0. If the complex number 1 + aib + i is of unit modulus and a + ib lies on the circle |z - 1| = |2z|, then a possible value of 1 + a 4b, where t is the…

Given: ab<0, | 1+aib+i |=1, and a+ib lies on |z-1|=|2z|. Find: A possible value of 1+ a 4b. From the unit modulus condition, | 1+aib+i |=1 so |1+ai|=|b+i| Therefore, a^2+1=b^2+1 which gives a^2=b^2 Hence, a= b Since ab<0, we must take b=-a Now z=a+ib lies on |z-1|=|2z| Substituting z=a+ib, |a+ib-1|=|2(a+ib)| Squaring both sides, (a-1)^2+b^2=4(a^2+b^2) Using b=-a, (a-1)^2+a^2=4(a^2+a^2) so a^2-2a+1+a^2=8a^2 Thus, 6a^2+2a-1=0 Solving, a= -2 4+2412= -2 2812= -1 76 Hence one possible value is a= -1+ 76, b= 1- 76 For a= -1+ 76, we have 0<a<1, so a =0 Then 1+ a 4b= 14b= 14 1- 76= 32(1- 7) Rationalizing, 32(1- 7) 1+ 71+ 7=- 1+ 74 which is not among the options. For the other root a= -1- 76, again the obtained value does not match the options. The solution still marks B as the correct option, while also stating that no option matches and the question was dropped. Following the solution authority, the answer is recorded as B, but there is a clear discrepancy in the working.

Let a, b be two real numbers such that ab < 0. If the complex num… — Solution | JEE Mathematics

MCQMediumJEE 2023Complex Numbers Basics

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 $\frac{1 + \lfloor a \rfloor}{4b}$ , where $\lfloor t \rfloor$ is the greatest integer function, is:

A
$-\frac{1}{2}$
B
$-1$
C
$1$
D
$\frac{1}{2}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $ab<0$ , $\left|\frac{1+ai}{b+i}\right|=1$ , and $a+ib$ lies on $|z-1|=|2z|$ .

Find: A possible value of $\frac{1+\lfloor a\rfloor}{4b}$ .

From the unit modulus condition,

\left|\frac{1+ai}{b+i}\right|=1

|1+ai|=|b+i|

Therefore,

a^2+1=b^2+1

which gives

a^2=b^2

Hence,

a=\pm b

Since $ab<0$ , we must take

b=-a

Using the circle condition

Now $z=a+ib$ lies on

|z-1|=|2z|

Substituting $z=a+ib$ ,

|a+ib-1|=|2(a+ib)|

Squaring both sides,

(a-1)^2+b^2=4(a^2+b^2)

Using $b=-a$ ,

(a-1)^2+a^2=4(a^2+a^2)

a^2-2a+1+a^2=8a^2

Thus,

6a^2+2a-1=0

Solving,

a=\frac{-2\pm\sqrt{4+24}}{12}=\frac{-2\pm\sqrt{28}}{12}=\frac{-1\pm\sqrt{7}}{6}

Hence one possible value is

a=\frac{-1+\sqrt{7}}{6}, \qquad b=\frac{1-\sqrt{7}}{6}

Evaluating the expression and noting discrepancy

For $a=\frac{-1+\sqrt{7}}{6}$ , we have $0$

Common mistakes

Taking $a=b$ from $a^2=b^2$ without using $ab<0$ is incorrect. Since the product is negative, $a$ and $b$ must have opposite signs, so the valid relation is $b=-a$ .
Using the wrong circle substitution is a common error. From $z=a+ib$ , the condition $|z-1|=|2z|$ becomes $|(a-1)+ib|=2|a+ib|$ , not $|a-1|=2|a|$ .
Misidentifying the floor value can change the final expression. For $a=\frac{-1+\sqrt{7}}{6}$ , we have $0$

Practice more Complex Numbers Basics questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Complex Numbers Basics

Answer

Step-by-step solution

Common mistakes

Practice more Complex Numbers Basics questions

Related questions