Square Root Shortcut for Complex Numbers in JEE Main – Ultimate Guide

Introduction

Finding $\sqrt{a+ib}$ traditionally requires solving simultaneous equations. This direct formula solves it in seconds, saving 2-3 minutes per problem.

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The Master Formula

For $z = a + ib$ with $b \neq 0$:

$\boxed{\sqrt{a+ib} = \pm\left( \sqrt{\frac{|z|+a}{2}} \ + \ i\cdot\frac{b}{|b|}\cdot\sqrt{\frac{|z|-a}{2}} \right)}$

Where:

• $|z| = \sqrt{a^2 + b^2}$ (modulus)
• $\frac{b}{|b|}$ = sign of $b$ (+1 if $b>0$, -1 if $b<0$)

Memory aid: "Plus-a for real part, minus-a for imaginary part, keep b's sign."

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Step-by-Step Algorithm

1. Compute modulus: $|z| = \sqrt{a^2+b^2}$
2. Real part magnitude: $R = \sqrt{\frac{|z|+a}{2}}$
3. Imaginary part magnitude: $I = \sqrt{\frac{|z|-a}{2}}$
4. Sign rule:
   • If $b > 0$: $\sqrt{z} = \pm(R + iI)$
   • If $b < 0$: $\sqrt{z} = \pm(R - iI)$

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Lightning-Fast Examples

Example 1: $\sqrt{3+4i}$

• $a=3, b=4$, $|z|=5$
• $R = \sqrt{\frac{5+3}{2}} = \sqrt{4} = 2$
• $I = \sqrt{\frac{5-3}{2}} = \sqrt{1} = 1$
• $b>0$ ⇒ $\sqrt{3+4i} = \pm(2+i)$ ✓

Time: 15 seconds

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Example 2: $\sqrt{8-6i}$

• $a=8, b=-6$, $|z|=10$
• $R = \sqrt{\frac{10+8}{2}} = \sqrt{9} = 3$
• $I = \sqrt{\frac{10-8}{2}} = \sqrt{1} = 1$
• $b<0$ ⇒ $\sqrt{8-6i} = \pm(3-i)$

Check: $(3-i)^2 = 9 - 6i + i^2 = 9-6i-1 = 8-6i$ ✓

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Example 3: $\sqrt{-5+12i}$

• $a=-5, b=12$, $|z|=13$
• $R = \sqrt{\frac{13+(-5)}{2}} = \sqrt{4} = 2$
• $I = \sqrt{\frac{13-(-5)}{2}} = \sqrt{9} = 3$
• $b>0$ ⇒ $\sqrt{-5+12i} = \pm(2+3i)$

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Special Cases (Instant Solutions)

Case 1: Purely Imaginary ($a=0$)

$\sqrt{ib} = \pm\sqrt{\frac{|b|}{2}}(1+i)$ if $b>0$

Example: $\sqrt{8i}$

• $|8i|=8$
• $\sqrt{8i} = \pm\sqrt{4}(1+i) = \pm2(1+i)$

Case 2: Negative Real ($b=0, a<0$)

$\sqrt{-r} = \pm i\sqrt{r}$

Example: $\sqrt{-9} = \pm3i$

Case 3: Perfect Square Modulus

When $|z|$ is integer, calculations are mental.

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JEE PYQs Solved Rapidly

PYQ 1: JEE Main 2019 Pattern

Find $\sqrt{7-24i}$

30-second solution:

• $a=7, b=-24$, $|z|=\sqrt{49+576}=25$
• $R=\sqrt{\frac{25+7}{2}}=\sqrt{16}=4$
• $I=\sqrt{\frac{25-7}{2}}=\sqrt{9}=3$
• $b<0$ ⇒ $\sqrt{7-24i}=\pm(4-3i)$

Answer: $\pm(4-3i)$

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PYQ 2: JEE Main 2021 Pattern

If $z^2 = -15 + 8i$, find $z$

Solution: $z = \sqrt{-15+8i}$

• $a=-15, b=8$, $|z|=\sqrt{225+64}=17$
• $R=\sqrt{\frac{17+(-15)}{2}}=\sqrt{1}=1$
• $I=\sqrt{\frac{17-(-15)}{2}}=\sqrt{16}=4$
• $b>0$ ⇒ $z=\pm(1+4i)$

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PYQ 3: Multiple Roots

Solve: $z^4 + 5z^2 + 6 = 0$

Solution: Let $w=z^2$: $w^2+5w+6=0$ $w=-2$ or $w=-3$

So $z^2=-2$ or $z^2=-3$ $z=\pm\sqrt{-2}=\pm i\sqrt{2}$ $z=\pm\sqrt{-3}=\pm i\sqrt{3}$

Answer: $\pm i\sqrt{2}, \pm i\sqrt{3}$

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Important Properties

1. Conjugate Property

$\sqrt{\overline{z}} = \overline{\sqrt{z}}$

If $\sqrt{a+ib}=x+iy$, then $\sqrt{a-ib}=x-iy$

2. Modulus Property

$|\sqrt{z}| = \sqrt{|z|}$

Example: $|\sqrt{3+4i}| = \sqrt{|3+4i|} = \sqrt{5}$

3. Product Property

$\sqrt{z_1} \cdot \sqrt{z_2} = \sqrt{z_1 z_2}$ (with careful branch selection)

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Common JEE Values (Memorize!)

*Not a nice integer combination

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Verification Shortcut

After finding $\sqrt{a+ib}=x+iy$, quickly check:

1. $x^2-y^2 = a$
2. $2xy = b$
3. $x^2+y^2 = |z|$

Example: For $\sqrt{3+4i}=2+i$:

• $2^2-1^2=4-1=3$ ✓
• $2(2)(1)=4$ ✓
• $2^2+1^2=5=|3+4i|$ ✓

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Polar Form Method (Alternative)

If $z=re^{i\theta}$, then $\sqrt{z}=\sqrt{r}e^{i\theta/2}$

Useful when $\theta$ is standard angle.

Example: $\sqrt{1+i\sqrt{3}}$

• $r=2$, $\theta=\pi/3$
• $\sqrt{z}=\sqrt{2}e^{i\pi/6}=\sqrt{2}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
• $=\sqrt{2}\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)=\frac{\sqrt{6}}{2}+i\frac{\sqrt{2}}{2}$

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Practice Problems

Problem 1: $\sqrt{-8+6i}$

Solution:

• $a=-8, b=6$, $|z|=10$
• $R=\sqrt{\frac{10+(-8)}{2}}=1$
• $I=\sqrt{\frac{10-(-8)}{2}}=3$
• $b>0$ ⇒ $\sqrt{-8+6i}=\pm(1+3i)$

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Problem 2: $\sqrt{40+42i}$

Solution:

• $a=40, b=42$, $|z|=\sqrt{1600+1764}=\sqrt{3364}=58$
• $R=\sqrt{\frac{58+40}{2}}=\sqrt{49}=7$
• $I=\sqrt{\frac{58-40}{2}}=\sqrt{9}=3$
• $b>0$ ⇒ $\sqrt{40+42i}=\pm(7+3i)$

Check: $(7+3i)^2=49+42i+9i^2=49+42i-9=40+42i$ ✓

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Problem 3: Find $z$ if $z^2 = \frac{1+i\sqrt{3}}{1-i\sqrt{3}}$

Solution: Simplify: $\frac{1+i\sqrt{3}}{1-i\sqrt{3}} = \frac{(1+i\sqrt{3})^2}{1+3} = \frac{-2+2i\sqrt{3}}{4} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$

So $z^2 = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$

This is $e^{i2\pi/3}$ in polar form ($r=1$, $\theta=2\pi/3$)

Thus $z = \pm e^{i\pi/3} = \pm\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$

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Quick Reference Chart

| Input | $|z|$ | Real Part | Imag Part | Sign | Result | |-------|------|-----------|-----------|------|--------| | $a+ib$ ($b>0$) | $\sqrt{a^2+b^2}$ | $\sqrt{\frac{\|z\|+a}{2}}$ | $\sqrt{\frac{\|z\|-a}{2}}$ | same | $\pm(R+iI)$ | | $a+ib$ ($b<0$) | $\sqrt{a^2+b^2}$ | $\sqrt{\frac{\|z\|+a}{2}}$ | $\sqrt{\frac{\|z\|-a}{2}}$ | opposite | $\pm(R-iI)$ |

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Exam Strategy

1. Check for special cases first (pure imaginary, negative real)
2. Compute modulus accurately (common source of error)
3. Apply formula systematically
4. Verify with quick check ($x^2-y^2=a$, $2xy=b$)
5. Remember $\pm$ gives two roots

Time target: 30-60 seconds per square root problem

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Common Mistakes to Avoid

1. Forgetting the $\pm$: Square roots give two values
2. Sign error: Mixing up when real/imag parts have same/opposite signs
3. Modulus error: $|a+ib|=\sqrt{a^2+b^2}$, not $\sqrt{a^2-b^2}$
4. Order error: $\sqrt{\frac{|z|+a}{2}}$ is real part, $\sqrt{\frac{|z|-a}{2}}$ is imaginary magnitude

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Final Tip: Practice with 5-10 examples to build speed. This formula alone can save 5+ minutes in JEE Main if complex number square roots appear.

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Related: Complex Algebra Shortcuts | Quick Formula Sheet