Complex Number Practice Questions

The square root of a negative number is defined using the imaginary unit \(i\), where \(i = \sqrt{-1}\). Separate the negative sign from the number under the radical:

\(\sqrt{-36} = \sqrt{-1 \times 36} = \sqrt{-1} \times \sqrt{36}\) \(\:= i \times 6 = 6i\)

Choice B drops the imaginary unit entirely. Remember: \(\sqrt{-36} \neq -6\) because \((-6)^2 = 36\), not −36. The result must involve \(i\).

The powers of \(i\) follow a repeating cycle of 4:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\) (then the cycle repeats)

To find \(i^{27}\), divide the exponent by 4 and use the remainder:

\(27 \div 4 = 6 \text{ remainder } 3\)

The remainder is 3, so \(i^{27} = i^3 = -i\).

To add complex numbers, combine the real parts and the imaginary parts separately:

\((4 + 3i) + (2 – 7i) = (4 + 2) + (3i – 7i)\)\(\:= 6 – 4i\)

This works just like combining like terms. The real parts (4 and 2) add to 6, and the imaginary parts (\(3i\) and \(-7i\)) add to \(-4i\).

Distribute the negative sign to both parts of the second complex number, then combine like terms:

\((5 + 2i) – (3 – 6i)\)\(\:= 5 + 2i – 3 + 6i\)\(\:= 2 + 8i\)

Choice A results from incorrectly computing \(-(- 6i)\) as \(-6i\) instead of \(+6i\). Remember: subtracting a negative imaginary part produces a positive imaginary part.

Multiply complex numbers using the FOIL method, just like binomials. Then replace \(i^2\) with −1:

\((3 + 2i)(4 – 5i)\)\(\:= 12 – 15i + 8i – 10i^2\)

Since \(i^2 = -1\):

\(= 12 – 7i – 10(-1)\)\(\:= 12 – 7i + 10\)\(\:= 22 – 7i\)

Choice B forgets to replace \(i^2\) with −1, leaving the \(-10i^2\) as −10 instead of +10.

The expressions \((2 + 3i)\) and \((2 – 3i)\) are complex conjugates. The product of conjugates follows the difference of squares pattern:

\((a + bi)(a – bi) = a^2 – (bi)^2 = a^2 – b^2i^2\)\(\:= a^2 + b^2\)

Substitute \(a = 2\) and \(b = 3\):

\((2)^2 + (3)^2 = 4 + 9 = 13\)

The product of complex conjugates is always a real number. Choice C stops before replacing \(i^2\) with −1.

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of \((1 – i)\) is \((1 + i)\):

\(\dfrac{4 + 2i}{1 – i} \cdot \dfrac{1 + i}{1 + i}\)

Multiply the numerator using FOIL:

\((4 + 2i)(1 + i)\)\(\:= 4 + 4i + 2i + 2i^2\)\(\:= 4 + 6i – 2 = 2 + 6i\)

Multiply the denominator (conjugate pairs produce a real number):

\((1 – i)(1 + i) = 1 + 1 = 2\)

Divide:

\(\dfrac{2 + 6i}{2} = 1 + 3i\)

The complex conjugate of a number \(a + bi\) is \(a – bi\). You change only the sign of the imaginary part (the real part stays the same).

For \(5 – 7i\), the conjugate is \(5 + 7i\).

Choice A negates both parts (which gives the negative of the conjugate, not the conjugate itself). Choice D swaps the real and imaginary parts, which is not what a conjugate does.

The modulus of a complex number \(a + bi\) is defined as:

\(|a + bi| = \sqrt{a^2 + b^2}\)

Substitute \(a = 3\) and \(b = 4\):

\(|3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16}\)\(\:= \sqrt{25} = 5\)

Geometrically, the modulus is the distance from the origin to the point \((3, 4)\) on the complex plane, which is a classic 3-4-5 right triangle.

Choice C gives \(a^2 + b^2 = 25\) without taking the square root.

Isolate \(x^2\):

\(x^2 = -16\)

Take the square root of both sides, remembering both the positive and negative roots:

\(x = \pm\sqrt{-16} = \pm\sqrt{16} \cdot \sqrt{-1} = \pm 4i\)

Choice A ignores the negative sign under the radical. Since \(x^2 = -16\) (not \(x^2 = 16\)), the solutions must be imaginary, not real.

Apply the quadratic formula with \(a = 1\), \(b = -4\), and \(c = 13\):

\(x = \dfrac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(13)}}{2(1)}\) \(\:= \dfrac{4 \pm \sqrt{16 – 52}}{2}\) \(\:= \dfrac{4 \pm \sqrt{-36}}{2}\)

The discriminant is negative, which means the solutions are complex. Simplify \(\sqrt{-36} = 6i\):

\(x = \dfrac{4 \pm 6i}{2} = 2 \pm 3i\)

The two solutions are \(x = 2 + 3i\) and \(x = 2 – 3i\). Notice they are complex conjugates — complex solutions to quadratics with real coefficients always come in conjugate pairs.

When multiplying square roots of negative numbers, rewrite each one using \(i\) first before multiplying:

\(\sqrt{-8} = i\sqrt{8} = 2i\sqrt{2}\), \(\:\sqrt{-2} = i\sqrt{2}\)

Now multiply:

\((2i\sqrt{2})(i\sqrt{2})\)\(\:= 2i^2 \cdot (\sqrt{2})^2\)\(\:= 2(-1)(2) = -4\)

It’s important to remember that you can’t use \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\) when \(a\) and \(b\) are both negative. Doing so would give \(\sqrt{(-8)(-2)} = \sqrt{16} = 4\) (choice A), which is incorrect. Always convert to \(i\) form first.

Expand using the formula \((a + b)^2 = a^2 + 2ab + b^2\):

\((1 + i)^2 = 1^2 + 2(1)(i) + i^2\)\(\:= 1 + 2i + (-1)\)\(\:= 2i\)

The real parts (1 and −1) cancel, leaving only the imaginary part. Choice A forgets the middle term \(2(1)(i) = 2i\) when squaring. Choice D computes \(i^2\) as \(+1\) instead of −1.

Replace each power of \(i\) with its value from the cycle:

• \(i = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

Now add them all:

\(i + (-1) + (-i) + 1 = 0\)

The terms cancel in pairs: \(i\) and \(-i\) cancel, and −1 and 1 cancel. In fact, any complete cycle of four consecutive powers of \(i\) will always sum to zero.

The conjugate of \(z = 3 + 2i\) is \(\bar{z} = 3 – 2i\). Multiply them using the conjugate product formula:

\(z \cdot \bar{z} = (3 + 2i)(3 – 2i) = 3^2 + 2^2\)\(\:= 9 + 4 = 13\)

Recall that the product of a complex number and its conjugate always equals \(a^2 + b^2\), which is a real number. In fact, \(z \cdot \bar{z} = |z|^2\), which is the square of the modulus.

Choices C and D would imply an imaginary part remains, but the product of conjugates always eliminates the imaginary part entirely.