Complex Numbers and Quadratic Equations MCQs

i is defined as the square root of –1, so i² = –1 by definition.

Since (–1) is the only real number whose square gives –1, i² cannot be 1, i, or –i.

Thus the value of i² is –1.

i^1 = i, i^2 = –1, i^3 = –i, i^4 = 1

Add the four terms: i + (–1) + (–i) + 1

i and –i cancel, –1 and +1 cancel, leaving 0

Hence the sum equals 0.

|z1z2| = √[(Re(z1)Re(z2)−Im(z1)Im(z2))^2+(Re(z1)Im(z2)+Im(z1)Re(z2))^2] which simplifies to √[(Re(z1)^2+Im(z1)^2)(Re(z2)^2+Im(z2)^2)] = √[|z1|^2·|z2|^2] = |z1|·|z2|. Hence the product of moduli equals the modulus of the product, giving |z1| |z2|.

The complex number i can be written as 0 + 1i, so its point in the Argand plane is (0, 1).

The argument is the angle made with the positive real axis, measured counter‑clockwise.

For the point (0, 1) this angle is 90°, i.e. π/2 radians.

Hence the principal argument of i is π/2.

Multiply numerator and denominator by the conjugate of the denominator: (1+i)(1+i)/(1‑i)(1+i)

The denominator becomes (1)^2‑(i)^2 = 1‑(‑1)=2.

The numerator expands to 1+2i+i^2 = 1+2i‑1 = 2i, so the fraction is (2i)/2 = i.

Hence the quotient equals i.

The conjugate of a complex number a+bi is a−bi.

Here the number is i = 0+1·i, so a = 0 and b = 1.

Replacing b by –b gives 0−1·i = −i.

Hence the conjugate of i is –i.

Find the modulus r = √(a²+b²) for z = a+ib with a=1, b=1.

Compute r = √(1²+1²) = √2.

Thus the modulus of 1+i is √2, which matches the given answer.

x^2+4=0 ⇒ x^2 = –4.

Write –4 as 4·(–1) = 4·i^2, so x = ±√(4·i^2) = ±2i.

Thus the two solutions are 2i and –2i, which matches the given answer.

i^2 = –1, so powers of i repeat every 4: i^1 = i, i^2 = –1, i^3 = –i, i^4 = 1.

Divide the exponent 7 by 4: 7 = 4·1 + 3, so i^7 = i^(4·1)·i^3 = (i^4)^1·i^3 = 1·(–i).

Thus i^7 = –i.

–i.

Subtract the real parts: 3 − 1 = 2.

Subtract the imaginary parts: 4i − (−2i) = 4i + 2i = 6i.

Combine them to get (2 + 6i), which matches the given answer.

Modulus of a real number is its distance from zero on the number line, i.e., absolute value.

For –7, |–7| = 7 because the distance from 0 to –7 is 7 units.

Hence the modulus is 7.

The complex number -1 lies on the negative real axis, so its angle with the positive real axis is measured from the origin to the point (-1,0).

The principal argument is taken in the interval (−π, π], and the angle for a point on the negative real axis is π radians.

Hence the principal argument of –1 is π.

Add the real parts: 2 + 4 = 6.

Add the imaginary parts: 3i + (–i) = 2i.

Combine them to get z = 6 + 2i, which matches the given answer.

Multiply the two complex numbers: (3+i)(1+i)=3·1+3i+ i·1+i²=3+3i+i−1.

Combine real parts: 3−1=2 and imaginary parts: 3i+i=4i.

Thus the product is 2+4i, whose imaginary part is 4.

Answer: 4

For any complex number \(z = a+bi\) its conjugate is \(\overline{z}=a-bi\).

Setting \(\overline{z}=z\) gives \(a-bi = a+bi\).

Thus \(b\) must be zero, leaving \(z=a\) which is a real number.

Hence \(z\) is purely real.

Real part =1, imaginary part =1, so tan θ = Im/Re =1/1 =1.

Hence principal argument θ = arctan 1 = π/4 (since both parts are positive, θ lies in the first quadrant).

Thus the argument of 1 + i is π/4.

Multiply using the conjugate: (2+3i)(2−3i)=2²−(3i)²

=4−9(i²) and i²=−1, so −9(i²)=−9(−1)=+9

Add the terms: 4+9=13, which matches the given answer. 13

Multiply numerator and denominator by the conjugate (2+3i) to rationalise.

(1)(2+3i) = 2+3i and (2-3i)(2+3i)=4+9=13.

Thus 1/(2‑3i) = (2+3i)/13, which matches the given answer.

Let z = a + ib, where a = Re(z) and b = Im(z).

Its conjugate \(\bar z = a - ib\).

Adding: \(z + \bar z = (a+ib)+(a-ib)=2a = 2\operatorname{Re}(z)\).

Hence the sum equals \(2\operatorname{Re}(z)\).

In the Argand plane the x‑coordinate gives the real part and the y‑coordinate gives the imaginary part of the complex number.

The point is \((-2,\,-3)\) so the real part is \(-2\) and the imaginary part is \(-3\).

Thus the complex number is \(-2-3i\).