Exams › JEE Main › Maths

If z is a complex number such that |z| 2, then the minimum value of |z + 1/2| is:

1. Is strictly greater than 5/2
2. Is strictly greater than 3/2 but less than 5/2
3. Is equal to 5/2
4. Lies in the interval (1, 2)

Correct answer: Lies in the interval (1, 2)

Solution

The minimum value of |z + 1/2| occurs when z is at its closest point to -1/2 on the complex plane, which is at a distance of 2 from the origin. This results in a minimum distance of 1, thus placing the minimum value of |z + 1/2| in the interval (1, 2).

Related JEE Main Maths questions

• Let $\alpha$ and $\beta$ denote the roots in $x$ of the equation $m^2(x^2-x)+2mx+3=0$. If $m_1$ and $m_2$ are the two values of $m$ such that $\alpha$ and $\beta$ satisfy the condition $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{4}{3}$, then find the value of $\frac{m_1^2}{m_2}-\frac{m_2^2}{m_1}$.
• Given $z_1=\sqrt{3}+i\sqrt{3}$ and $z_2=\sqrt{3}+i$, determine the quadrant in which the complex number $\frac{z_1}{z_2}$ lies.
• For the quadratic equation $2(1+i)x^2-4(2-i)x-(5+3i)=0$, the root with the larger absolute value is
• Find the value of $\dfrac{(\cos\theta+i\sin\theta)^4}{(\cos\theta-i\sin\theta)^3}$.
• How many complex numbers $z$ satisfy the equation $z^3+\dfrac{3|z|^2}{z}=0$, given that $|z|=\sqrt{3}$?
• Let $z=x+iy$ be a variable complex number. If $\arg\!\left(\dfrac{z-1}{z+1}\right)=\dfrac{\pi}{4}$, then which relation is satisfied?

⚔️ Practice JEE Main Maths free + battle 1v1 →