Let z Be a Complex Number Such That |z + 2| = 1 and Im((z+1)/(z+2)) = 1/5. Then the Value of |Re(z + 2)| Is

JEE Main 2024. Previous Year Question.

Problem

Let $z$ be a complex number such that $|z+2|=1$ and $\operatorname{Im}\left(\dfrac{z+1}{z+2}\right)=\dfrac15$. Then the value of $|\operatorname{Re}(z+2)|$ is:

Key insight. Since $|z+2|=1$ means $(x+2)^2+y^2=1$, the denominator of $\dfrac{z+1}{z+2}$, once rationalised, has modulus-squared exactly $1$ — so it simply disappears from the imaginary part. That turns a messy complex fraction into one clean equation for $y$ alone.

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Approach

Write $z=x+iy$ and translate $|z+2|=1$ into $(x+2)^2+y^2=1$. Rationalise $\dfrac{z+1}{z+2}$ by multiplying by the conjugate of the denominator, extract its imaginary part, and use the circle equation to simplify the denominator to $1$. Solve for $y$, then use the circle equation again to find $|x+2|$.

Solution

Step 1 — Translate the modulus condition

$$|z+2|=1 \implies (x+2)^2+y^2=1$$

Step 2 — Rationalise (z+1)/(z+2)

$$\frac{z+1}{z+2}=\frac{(x+1)+iy}{(x+2)+iy}=\frac{[(x+1)+iy][(x+2)-iy]}{(x+2)^2+y^2}$$

Step 3 — Extract the imaginary part of the numerator

$$\text{Im of numerator} = y(x+2)-y(x+1) = y[(x+2)-(x+1)] = y$$

Step 4 — Use the denominator’s value from Step 1

$$\operatorname{Im}\left(\frac{z+1}{z+2}\right) = \frac{y}{(x+2)^2+y^2} = \frac{y}{1} = y$$

Step 5 — Apply the given condition

$$y = \frac15$$

Step 6 — Find |x + 2| using the circle equation

$$(x+2)^2 = 1-y^2 = 1-\frac{1}{25} = \frac{24}{25}$$

$$|x+2| = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt6}{5}$$

Step 7 — Identify the answer

Since $\operatorname{Re}(z+2)=x+2$:

$$|\operatorname{Re}(z+2)| = \frac{2\sqrt6}{5}$$

Answer

Common mistakes

• Rationalising the whole fraction and computing the real part too, when only the imaginary part is needed. The question’s condition is entirely on $\operatorname{Im}\left(\frac{z+1}{z+2}\right)$ — computing the real part as well wastes time and adds room for error.
• Forgetting that the denominator’s modulus-squared equals 1 by the given circle condition. Without spotting this, the imaginary part looks like a genuinely two-variable expression; recognising $(x+2)^2+y^2=1$ collapses it to just $y$.

Practise next

• Let $z$ satisfy $|z-1|=1$ and $\operatorname{Im}\left(\dfrac{z}{z-1}\right)=\dfrac13$. Find $|\operatorname{Re}(z-1)|$, using the same rationalise-and-substitute approach.