Complex Numbers, Powers of i and Square Roots of Negatives

Why Invent New Numbers?

The equation $x^2 + 1 = 0$ demands a number whose square is $-1$. No real number qualifies — squares of reals are never negative. Faced with this, mathematics does what it has always done: build a bigger number system. Naturals grew into integers (to solve $x + 5 = 0$), integers into rationals (to solve $2x = 1$), rationals into reals (to solve $x^2 = 2$) — and now reals grow into complex numbers, where $x^2 + 1 = 0$ finally has roots.

Key Point (Definition): Denote $\sqrt{-1}$ by the symbol i, so that $i^2 = -1$. A number of the form $a + ib$, where a and b are real, is a complex number. For $z = a + ib$: a is the real part (Re z) and b is the imaginary part (Im z).

Examples: $2 + i3$, $(-1) + i\sqrt{3}$, $4 + i\left(\frac{-1}{11}\right)$. For $z = 2 + i5$: Re z = 2 and Im z = 5. Every real number is a complex number with imaginary part 0 (e.g. $7 = 7 + i0$), and numbers like $i5 = 0 + i5$ are called purely imaginary.

Equality of complex numbers

$a + ib = c + id$ exactly when $a = c$ AND $b = d$ — real parts match and imaginary parts match. One complex equation therefore packs TWO real equations, a fact used constantly to solve for unknowns.

[Board Important] In this chapter the roadmap is: definition and powers of i (this section) → arithmetic (Section 2) → modulus and conjugate (Section 3) → the Argand plane (Section 4). JEE extras (polar form, quadratic equations, cube roots of unity) live in the JEE Corner.

The Powers of i — a Four-Cycle

Multiplying by i over and over produces a repeating wheel:

$$i^1 = i, \quad i^2 = -1, \quad i^3 = i^2 \cdot i = -i, \quad i^4 = (i^2)^2 = 1$$

and then it repeats: $i^5 = i$, $i^6 = -1$, and so on. Negative powers run the wheel backwards: $i^{-1} = \frac{1}{i} = \frac{i}{i^2} = -i$, $i^{-2} = -1$, $i^{-3} = i$, $i^{-4} = 1$.

Key Point: For any integer k: $i^{4k} = 1$, $i^{4k+1} = i$, $i^{4k+2} = -1$, $i^{4k+3} = -i$. To evaluate any power of i, divide the exponent by 4 — the remainder decides everything.

Worked instances: $i^{35} = i^{4 \times 8 + 3} = -i$; $\quad i^{-35} = \frac{1}{i^{35}} = \frac{1}{-i} = \frac{i}{-i^2} = i$; $\quad i^9 + i^{19} = i - i \cdot i^2 \cdot ... = i^{8+1} + i^{16+3} = i + (-i) = 0$.

[JEE Tip] For big negative exponents, add a convenient multiple of 4 instead of dividing: $i^{-39} = i^{-39} \cdot i^{40} = i^1 = i$. Multiplying by $i^{4k} = 1$ is always free.

Square Roots of Negative Real Numbers

Since $i^2 = -1$ and $(-i)^2 = -1$, the square roots of $-1$ are $i$ and $-i$. By convention, the symbol $\sqrt{-1}$ means i only. More generally, for a positive real a:

$$\sqrt{-a} = i\sqrt{a}$$

So $\sqrt{-3} = \sqrt{3}\,i$ and $\sqrt{-4} = 2i$ (while the equation $x^2 = -3$ has BOTH solutions $\pm\sqrt{3}i$).

The $\sqrt{a}\sqrt{b}$ trap

The familiar rule $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ holds when at least one of a, b is non-negative — but FAILS when both are negative. The famous contradiction if we ignore this:

$$-1 = i^2 = \sqrt{-1}\,\sqrt{-1} \overset{?}{=} \sqrt{(-1)(-1)} = \sqrt{1} = 1$$

The step marked ? is illegal. Correct procedure: convert each root to $i\sqrt{a}$ form FIRST, then multiply: $\sqrt{-2}\,\sqrt{-3} = (i\sqrt{2})(i\sqrt{3}) = i^2\sqrt{6} = -\sqrt{6}$.

[JEE Tip] This trap is a direct MCQ generator: "$\sqrt{-9} \times \sqrt{-16}$ = ?" Answer: $(3i)(4i) = -12$, NOT +12. Convert to i-form before every multiplication involving two negative radicands.

Solved Examples

Example 1: Equality of complex numbers

If $4x + i(3x - y) = 3 + i(-6)$, where x and y are real numbers, find x and y.

Solution:

Step 1 — Equate the real parts. $4x = 3$, so $x = \frac{3}{4}$.

Step 2 — Equate the imaginary parts. $3x - y = -6$.

Step 3 — Substitute x and solve for y. $y = 3x + 6 = 3 \cdot \frac{3}{4} + 6 = \frac{9}{4} + \frac{24}{4} = \frac{33}{4}$.

Step 4 — Check. $4 \cdot \frac{3}{4} = 3$ ✓ and $\frac{9}{4} - \frac{33}{4} = -6$ ✓.

Final Answer: $x = \frac{3}{4}$, $y = \frac{33}{4}$.

Takeaway: One complex equation = two real equations. Split and solve simultaneously.

Example 2: Identifying parts

Write the real and imaginary parts of: (i) $z = -7 + 2i$ (ii) $z = -i$ (iii) $z = 5$ (iv) $z = \frac{1 - i}{3}$.

Solution:

Step 1 — (i) Read off directly. Re z = $-7$, Im z = 2.

Step 2 — (ii) Write in standard form. $z = 0 + (-1)i$: Re z = 0, Im z = $-1$ — purely imaginary.

Step 3 — (iii) Standard form again. $z = 5 + 0i$: Re z = 5, Im z = 0 — a real number.

Step 4 — (iv) Split the fraction. $z = \frac{1}{3} - \frac{1}{3}i$: Re z = $\frac{1}{3}$, Im z = $-\frac{1}{3}$.

Takeaway: Im z is the COEFFICIENT of i — a real number, never including the i itself.

Example 3: Powers of i, the remainder method

Evaluate: (i) $i^{35}$ (ii) $i^{102}$ (iii) $i^{-39}$ (iv) $i^{2026}$.

Solution:

Step 1 — (i) Divide the exponent by 4. $35 = 4(8) + 3$, remainder 3: $i^{35} = i^3 = -i$.

Step 2 — (ii) Same method. $102 = 4(25) + 2$, remainder 2: $i^{102} = i^2 = -1$.

Step 3 — (iii) Negative exponent: add a multiple of 4. $i^{-39} = i^{-39} \cdot i^{40} = i^1 = i$ (multiplying by $i^{40} = 1$ is free).

Step 4 — (iv) One more remainder. $2026 = 4(506) + 2$: $i^{2026} = i^2 = -1$.

Takeaway: Remainder 0, 1, 2, 3 ↔ value 1, i, $-1$, $-i$. Nothing else to memorise.

Example 4: A sum of powers

Express $i^9 + i^{19}$ in the form $a + ib$.

Solution:

Step 1 — Reduce each power by its remainder. $9 = 4(2) + 1$ gives $i^9 = i$; and $19 = 4(4) + 3$ gives $i^{19} = i^3 = -i$.

Step 2 — Add. $i + (-i) = 0$.

Step 3 — Standard form. $0 = 0 + i0$.

Takeaway: Powers whose remainders are 1 and 3 cancel in pairs.

Example 5: i to a negative power

Express $i^{-39}$ in the form $a + ib$.

Solution:

Step 1 — Multiply by a free factor of 1. $i^{40} = (i^4)^{10} = 1$, so $i^{-39} = i^{-39} \cdot i^{40} = i^{1}$.

Step 2 — Standard form. $i = 0 + i(1)$.

Takeaway: The add-a-multiple-of-4 move beats computing reciprocals every time.

Example 6: Products with roots of negatives

Evaluate: (i) $\sqrt{-16} + 3\sqrt{-25}$ (ii) $\sqrt{-4} \times \sqrt{-9}$ (iii) $(\sqrt{-2})^2$.

Solution:

Step 1 — (i) Convert each radical to i-form, then add. $\sqrt{-16} = 4i$ and $\sqrt{-25} = 5i$: $4i + 3(5i) = 19i$.

Step 2 — (ii) Convert BEFORE multiplying. $(2i)(3i) = 6i^2 = -6$ — not +6; the rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails for two negative radicands.

Step 3 — (iii) Square the i-form. $(i\sqrt{2})^2 = i^2 \cdot 2 = -2$.

Takeaway: Every negative radicand becomes $i\sqrt{a}$ before any other algebra happens.

Example 7: The i-cycle in an expression

Evaluate $\left[i^{18} + \left(\frac{1}{i}\right)^{25}\right]^3$.

Solution:

Step 1 — Reduce the first term. $18 = 4(4) + 2$: $i^{18} = i^2 = -1$.

Step 2 — Reduce the second term. $\left(\frac{1}{i}\right)^{25} = i^{-25} = i^{-25} \cdot i^{28} = i^3 = -i$.

Step 3 — Set up the cube. The bracket is $-1 - i = -(1 + i)$, so the cube is $-(1 + i)^3$.

Step 4 — Expand with the binomial identity. $(1 + i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i$.

Step 5 — Attach the minus. $-(-2 + 2i) = 2 - 2i$.

Final Answer: $2 - 2i$.

Takeaway: Reduce every power first, then expand the small cube with the binomial identity.

Example 8: Solving with equality

Find real x, y if $(x + y) + i(x - y) = 6 + i2$.

Solution:

Step 1 — Split into two real equations. $x + y = 6$ and $x - y = 2$.

Step 2 — Add the equations. $2x = 8$, so $x = 4$.

Step 3 — Back-substitute. $y = 6 - 4 = 2$.

Step 4 — Check. $(4 + 2) + i(4 - 2) = 6 + 2i$ ✓.

Takeaway: Complex equality hands you a friendly linear system.

Example 9: An equation forcing complex roots

Solve $x^2 + 9 = 0$ over the complex numbers.

Solution:

Step 1 — Isolate the square. $x^2 = -9$.

Step 2 — Take both square roots. $x = \pm\sqrt{-9} = \pm 3i$.

Step 3 — Check one root. $(3i)^2 = 9i^2 = -9$ ✓.

Takeaway: Equations dead in $\mathbb{R}$ come alive in $\mathbb{C}$ — and roots arrive in $\pm$ pairs.

Example 10: Sum of four consecutive powers

Show that $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$ for every integer n.

Solution:

Step 1 — Factor out the smallest power. $i^n(1 + i + i^2 + i^3)$.

Step 2 — Evaluate the bracket. $1 + i + (-1) + (-i) = 0$.

Step 3 — Conclude. Anything times 0 is 0, for every n. ∎

Takeaway: Any four consecutive powers of i cancel — the tool for monster sums like $\sum_{n=1}^{2025} i^n$ (group in fours; the leftover terms decide).

Example 11: A long power sum

Evaluate $i + i^2 + i^3 + \ldots + i^{102}$.

Solution:

Step 1 — Group in blocks of four. Terms $i^1$ through $i^{100}$ form 25 complete blocks of four consecutive powers — each block sums to 0.

Step 2 — Handle the leftovers. Only $i^{101}$ and $i^{102}$ remain: $101 = 4(25) + 1$ gives $i^{101} = i$; $102 = 4(25) + 2$ gives $i^{102} = -1$.

Step 3 — Add. $i + (-1) = -1 + i$.

Final Answer: $-1 + i$.

Takeaway: Long geometric strings of i collapse to at most three leftover terms.

Example 12: Purely real or purely imaginary?

For what real values of x is $z = (x^2 - 4) + i(x - 2)$ (i) purely imaginary? (ii) real?

Solution:

Step 1 — (i) Write both conditions. Purely imaginary needs Re z = 0 AND Im z $\neq$ 0.

Step 2 — Solve the real-part condition. $x^2 - 4 = 0$ gives $x = \pm 2$.

Step 3 — Test each against the imaginary-part condition. At x = 2: Im z = 0 — rejected (z = 0 is not purely imaginary). At x = $-2$: Im z = $-4 \neq 0$ ✓. So x = $-2$.

Step 4 — (ii) Real needs Im z = 0. $x - 2 = 0$: x = 2 (then z = 0, which counts as real).

Takeaway: "Purely imaginary" has TWO conditions — zero real part AND non-zero imaginary part. Check both.