Q.If z1=2+3i and z2=4−5i, find
Concept understanding — Algebra of Complex Numbers
A complex number extends the reals by introducing the imaginary unit i, defined by i² = −1. Every complex number is written in the standard form z = a + ib, where a = Re(z) is the real part and b = Im(z) is the imaginary part — and crucially both a and b are real numbers (Im(z) is the coefficient b, not ib). This concept is the arithmetic of these numbers: adding, multiplying, dividing, and reading off when the result is real or imaginary.
1 — Powers of i. Because i² = −1, the powers cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, and then it repeats. To reduce any iⁿ, divide the exponent by 4 and keep the remainder r: iⁿ = iʳ. So i²⁰²⁴ = i⁰ = 1 (2024 is a multiple of 4) and i⁵⁷⁴ = i² = −1 (574 = 4·143 + 2). Negative powers use the same wheel: i⁻¹ = 1/i = −i, i⁻² = −1. A neat consequence: any four consecutive powers of i sum to 0 (iⁿ + iⁿ⁺¹ + iⁿ⁺² + iⁿ⁺³ = 0), which collapses long sums like i + i² + … + iⁿ to at most three surviving terms.
2 — Square roots of negative reals. For a > 0, √(−a) = i√a (so √(−16) = 4i). Danger: the rule √x · √y = √(xy) is valid only when at least one of x, y is non-negative. With two negatives you must convert to i-form first: √(−4) · √(−9) = (2i)(3i) = 6i² = −6, not √36 = 6. Mishandling this sign is one of the most common slips in the chapter.
3 — Addition, subtraction, multiplication. Add/subtract componentwise: (a + ib) ± (c + id) = (a ± c) + i(b ± d). Multiply like binomials and replace i² = −1: (a + ib)(c + id) = (ac − bd) + i(ad + bc). Useful special products: (1 + i)² = 2i, (1 − i)² = −2i, and a number times its conjugate gives a real number: (a + ib)(a − ib) = a² + b².
4 — The conjugate and division. The conjugate of z = a + ib is z̄ = a − ib (flip the sign of the imaginary part). To write a quotient in standard form, multiply top and bottom by the conjugate of the denominator — this makes the denominator real:
(a + ib)/(c + id) = (a + ib)(c − id) / (c² + d²).
For example (2 + 3i)/(1 − i) = (2 + 3i)(1 + i)/2 = (−1 + 5i)/2. The multiplicative inverse is the special case z⁻¹ = z̄ / (a² + b²) (dividing 1 by z). The additive inverse is simply −z = −a − ib.
5 — Powers of (1 ± i). These recur constantly. Build them from (1 + i)² = 2i: (1 + i)⁴ = (2i)² = −4, (1 + i)⁸ = 16, and (1 + i)/(1 − i) = i (so ((1 + i)/(1 − i))ⁿ = iⁿ). Reducing a high power to a power of 2i or of i beats multiplying it out — and stays inside pure algebra (no polar form needed here).
6 — Equality, and "purely real / purely imaginary". Two complex numbers are equal iff their real parts are equal and their imaginary parts are equal — one complex equation gives two real equations, which is how you solve for unknown reals x, y. A number is purely real when its imaginary part is 0, and purely imaginary when its real part is 0 (and it is non-zero). To find a parameter making an expression real or imaginary, first reduce the expression to a + ib form, then set the appropriate part to zero. A classic trap: "real" means Im = 0, not Re = 0 — do not confuse the two.
7 — Solving for an exponent. Some questions ask for the least positive integer n meeting a condition (e.g. iⁿ = 1, or (1 + i)ⁿ real / purely imaginary, or (1 + i)ⁿ = (1 − i)ⁿ). Translate the condition into a statement about the period: iⁿ = 1 needs n a multiple of 4, so the least is 4; (1 + i)ⁿ = (2i)^(n/2) is real when n/2 is even, i.e. n = 4. Watch off-by-one errors on the period.
How this concept is examined. JEE Main tests: reducing a power iⁿ (or a sum of powers of i); simplifying an expression to a + ib; multiplying/expanding; dividing by rationalising with the conjugate; reading off Re or Im of a quotient; solving an equality of complex numbers for real unknowns; deciding a parameter for a purely real or purely imaginary result; computing z⁻¹; evaluating powers of (1 ± i); the √(−a)·√(−b) pitfall; and finding the least n for a power condition. The habits that prevent most errors: replace i² = −1 immediately, keep Im(z) as the real coefficient b, convert every √(negative) to i√· before multiplying, and always rationalise a quotient before judging whether it is real or imaginary.
Add or subtract the real parts together and the imaginary parts together.
(i) 6−2i; (ii) −2+8i.
(i) (2+4)+(3−5)i=6−2i. (ii) (2−4)+(3−(−5))i=−2+8i.
(i) z1+z2=6−2i; (ii) z1−z2=−2+8i
- Sum. Combine like parts:
z1+z2=(2+3i)+(4−5i)=(2+4)+(3+(−5))i=6−2i.
- Difference. Subtract the second from the first, watching the sign on −5i:
Check (independent recomputation): adding back (z1−z2)+z2=(−2+8i)+(4−5i)=2+3i=z1, confirming (ii); and (z1+z2)−z2=(6−2i)−(4−5i)=2+3i=z1, confirming (i).
z1−z2=(2+3i)−(4−5i)=(2−4)+(3−(−5))i=−2+8i.
✓Final answer(i) z1+z2=6−2i; (ii) z1−z2=−2+8i
Mishandling the double negative in subtraction: 3−(−5)=8, not −2. Keep the subtraction sign attached to both parts of z2.
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