Complex Numbers Problems with Solutions - Grade 12

a) \( (3 + 2i) - (8 - 5i) \)

Distribute the negative sign and group the real and imaginary parts together:

\[ = 3 + 2i - 8 + 5i \]

\[ = (3 - 8) + (2 + 5)i = \mathbf{-5 + 7i} \]

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b) \( (4 - 2i)(1 - 5i) \)

Use the FOIL method to expand the binomials, and remember that \( i^2 = -1 \):

\[ = 4(1) + 4(-5i) - 2i(1) - 2i(-5i) \]

\[ = 4 - 20i - 2i + 10i^2 \]

\[ = 4 - 22i + 10(-1) \]

\[ = (4 - 10) - 22i = \mathbf{-6 - 22i} \]

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c) \( \dfrac{-2 - 4i}{i} \)

To remove the imaginary unit from the denominator, multiply the numerator and denominator by \( -i \) (the conjugate of \( i \)):

\[ = \dfrac{(-2 - 4i)(-i)}{(i)(-i)} = \dfrac{2i + 4i^2}{-i^2} \]

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

\[ = \dfrac{2i + 4(-1)}{-(-1)} = \dfrac{-4 + 2i}{1} = \mathbf{-4 + 2i} \]

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d) \( \dfrac{-3 + 2i}{3 - 6i} \)

Multiply the numerator and denominator by the complex conjugate of the denominator, which is \( 3 + 6i \):

\[ = \dfrac{(-3 + 2i)(3 + 6i)}{(3 - 6i)(3 + 6i)} \]

Expand both the numerator and the denominator:

Numerator: \( -9 - 18i + 6i + 12i^2 = -9 - 12i - 12 = -21 - 12i \)

Denominator: \( 3^2 - (6i)^2 = 9 - 36(-1) = 9 + 36 = 45 \)

Divide both terms by 45 and simplify the fractions:

\[ = \dfrac{-21 - 12i}{45} = \dfrac{-21}{45} - \dfrac{12}{45}i = \mathbf{-\dfrac{7}{15} - \dfrac{4}{15}i} \]

First, we isolate the complex number \( x + yi \) by multiplying both sides of the given equation by \( i \):

\[ \dfrac{x + yi}{i} \cdot i = (7 + 9i) \cdot i \]

\[ x + yi = 7i + 9i^2 \]

Substitute \( i^2 = -1 \) and write in standard form:

\[ x + yi = -9 + 7i \]

The expression we need to evaluate is \( (x + yi)(x - yi) \). Notice that \( (x - yi) \) is simply the complex conjugate of \( (x + yi) \). Therefore:

\[ (x + yi)(x - yi) = (-9 + 7i)(-9 - 7i) \]

This is a difference of squares: \( (a + bi)(a - bi) = a^2 + b^2 \). Applying this rule:

\[ = (-9)^2 + (7)^2 = 81 + 49 = \mathbf{130} \]

Let the unknown complex number be \( z = a + bi \), where \( a \) and \( b \) are real numbers. Its complex conjugate is therefore \( z' = a - bi \).

Substitute \( z \) and \( z' \) into the given equation:

\[ (a + bi) + 3(a - bi) = 5 - 6i \]

Expand and group the real terms and imaginary terms on the left side:

\[ a + bi + 3a - 3bi = 5 - 6i \]

\[ (a + 3a) + (b - 3b)i = 5 - 6i \]

\[ 4a - 2bi = 5 - 6i \]

Two complex numbers are equal if and only if their corresponding real and imaginary parts are strictly equal. By equating the parts, we form a system of two simple linear equations:

Real parts: \( 4a = 5 \implies a = \dfrac{5}{4} \)

Imaginary parts: \( -2b = -6 \implies b = 3 \)

Finally, substitute \( a \) and \( b \) back into our original definition of \( z \):

\[ \mathbf{z = \dfrac{5}{4} + 3i} \]

Let \( z = a + bi \), which makes its conjugate \( z' = a - bi \).

The product of a complex number and its conjugate always yields the sum of the squares of its components:

\[ z z' = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2(-1) = a^2 + b^2 \]

We are given that \( zz' = 25 \), so our first equation is:

\[ a^2 + b^2 = 25 \]

We are also given the linear equation \( a + b = 7 \). We can solve this system by substitution. Isolate \( b \) in the linear equation:

\[ b = 7 - a \]

Substitute this expression for \( b \) into the quadratic equation:

\[ a^2 + (7 - a)^2 = 25 \]

Expand the perfect square binomial:

\[ a^2 + (49 - 14a + a^2) = 25 \]

Combine like terms to form a standard quadratic equation (\( Ax^2 + Bx + C = 0 \)):

\[ 2a^2 - 14a + 24 = 0 \]

Divide the entire equation by 2 to simplify:

\[ a^2 - 7a + 12 = 0 \]

Factor the quadratic equation:

\[ (a - 4)(a - 3) = 0 \]

This gives two possible values for \( a \). We use \( b = 7 - a \) to find the corresponding values for \( b \):

• If \( a = 4 \), then \( b = 7 - 4 = 3 \). This gives the complex number \( z_1 = 4 + 3i \).
• If \( a = 3 \), then \( b = 7 - 3 = 4 \). This gives the complex number \( z_2 = 3 + 4i \).

Therefore, the complex numbers that satisfy both conditions are:

\[ \mathbf{z = 4 + 3i \quad \text{and} \quad z = 3 + 4i} \]

Method 1: Direct Substitution (as shown below)

a) Since \( x = 2 + 4i \) is a root, substituting it into the equation must yield zero:

\[ (2 + 4i)^2 + b(2 + 4i) + c = 0 \]

Expand the squared binomial and distribute the \( b \):

\[ (4 + 16i + 16i^2) + 2b + 4bi + c = 0 \]

Substitute \( i^2 = -1 \) and simplify:

\[ 4 + 16i - 16 + 2b + 4bi + c = 0 \]

\[ -12 + 16i + 2b + 4bi + c = 0 \]

Group the real and imaginary parts together:

\[ (-12 + 2b + c) + (16 + 4b)i = 0 \]

For a complex number to equal 0, both its real and imaginary parts must be 0:

Imaginary part: \( 16 + 4b = 0 \implies 4b = -16 \implies \mathbf{b = -4} \)

Real part: \( -12 + 2b + c = 0 \). Substitute \( b = -4 \):

\[ -12 + 2(-4) + c = 0 \implies -12 - 8 + c = 0 \implies \mathbf{c = 20} \]

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b) Because the polynomial coefficients (\( 1, b, c \)) are all real numbers, the Complex Conjugate Root Theorem states that the roots must occur in conjugate pairs. Therefore, the second root is \( 2 - 4i \).

Check: Substitute \( x = 2 - 4i \), \( b = -4 \), and \( c = 20 \) into the equation:

\[ (2 - 4i)^2 - 4(2 - 4i) + 20 \]

\[ = (4 - 16i + 16i^2) - 8 + 16i + 20 \]

\[ = (4 - 16i - 16) - 8 + 16i + 20 \]

\[ = -12 - 16i - 8 + 16i + 20 = (-12 - 8 + 20) + (-16 + 16)i = 0 \]

(Alternative Method: Use Vieta's Formulas. The sum of the roots is \( -b \), and the product is \( c \). Given roots \( 2+4i \) and \( 2-4i \), Sum = \( 4 \implies b = -4 \). Product = \( (2)^2 + (4)^2 = 20 \implies c = 20 \)).

Let \( z = a + bi \), where \( a \) and \( b \) are real numbers.

Substitute this into the given equation:

\[ (a + bi)^2 = -1 + 2\sqrt{6}i \]

Expand the left side:

\[ a^2 + 2abi + b^2i^2 = -1 + 2\sqrt{6}i \]

\[ a^2 - b^2 + 2abi = -1 + 2\sqrt{6}i \]

Equate the real and imaginary parts to form a system of equations:

1) \( a^2 - b^2 = -1 \)

2) \( 2ab = 2\sqrt{6} \)

From equation 2, isolate \( b \):

\[ ab = \sqrt{6} \implies b = \dfrac{\sqrt{6}}{a} \]

Substitute this expression for \( b \) into equation 1:

\[ a^2 - \left( \dfrac{\sqrt{6}}{a} \right)^2 = -1 \]

\[ a^2 - \dfrac{6}{a^2} = -1 \]

Multiply the entire equation by \( a^2 \) to clear the denominator:

\[ a^4 - 6 = -a^2 \]

\[ a^4 + a^2 - 6 = 0 \]

This is a quadratic equation in terms of \( a^2 \). We can factor it as:

\[ (a^2 + 3)(a^2 - 2) = 0 \]

This gives \( a^2 = -3 \) or \( a^2 = 2 \). Since \( a \) must be a real number, \( a^2 \) cannot be negative. Therefore, we discard \( a^2 = -3 \).

Solving \( a^2 = 2 \) gives two real roots: \( a = \sqrt{2} \) and \( a = -\sqrt{2} \).

Now, use \( b = \dfrac{\sqrt{6}}{a} \) to find the corresponding values for \( b \):

• If \( a = \sqrt{2} \), then \( b = \dfrac{\sqrt{6}}{\sqrt{2}} = \sqrt{\dfrac{6}{2}} = \sqrt{3} \).
• If \( a = -\sqrt{2} \), then \( b = \dfrac{\sqrt{6}}{-\sqrt{2}} = -\sqrt{3} \).

Combining these into our complex number \( z = a + bi \), we get two solutions:

\[ \mathbf{z_1 = \sqrt{2} + \sqrt{3}i \quad \text{and} \quad z_2 = -\sqrt{2} - \sqrt{3}i} \]

Let \( z = a + bi \) and \( z' = a - bi \), where \( a \) and \( b \) are real numbers.

Substitute these expressions into the given equation:

\[ (4 + 2i)(a + bi) + (8 - 2i)(a - bi) = -2 + 10i \]

Carefully expand both sets of products using FOIL:

First product: \( 4a + 4bi + 2ai + 2bi^2 = 4a - 2b + (2a + 4b)i \)

Second product: \( 8a - 8bi - 2ai + 2bi^2 = 8a - 2b + (-2a - 8b)i \)

Add the expanded expressions together:

\[ (4a - 2b + 8a - 2b) + (2a + 4b - 2a - 8b)i = -2 + 10i \]

Simplify by grouping like terms:

\[ (12a - 4b) + (-4b)i = -2 + 10i \]

Equate the real and imaginary parts to form a system of equations:

Imaginary part: \( -4b = 10 \implies \mathbf{b = -\dfrac{5}{2}} \)

Real part: \( 12a - 4b = -2 \). Substitute the value of \( b \):

\[ 12a - 4\left(-\dfrac{5}{2}\right) = -2 \]

\[ 12a + 10 = -2 \]

\[ 12a = -12 \implies \mathbf{a = -1} \]

Therefore, the complex number \( z \) is:

\[ \mathbf{z = -1 - \dfrac{5}{2}i} \]

Because all the coefficients in the cubic polynomial (\( 1, 6, 61, 106 \)) are real numbers, the Complex Conjugate Root Theorem guarantees that if \( z_1 = -2 + 7i \) is a root, its conjugate \( z_2 = -2 - 7i \) must also be a root.

If we know two roots, we can create the quadratic factor that corresponds to them by multiplying \( (z - z_1)(z - z_2) \):

\[ (z - (-2 + 7i))(z - (-2 - 7i)) \]

To multiply this easily, regroup the inner terms to create a difference of squares:

\[ = ((z + 2) - 7i)((z + 2) + 7i) \]

\[ = (z + 2)^2 - (7i)^2 \]

\[ = (z^2 + 4z + 4) - 49(-1) \]

\[ = z^2 + 4z + 53 \]

This quadratic expression must be a factor of our original cubic polynomial. Let the remaining real root be \( r \). The cubic polynomial can be factored as:

\[ z^3 + 6z^2 + 61z + 106 = (z^2 + 4z + 53)(z - r) \]

We can find the missing linear factor \( (z - r) \) by performing polynomial long division, dividing the cubic polynomial by the quadratic factor. Alternatively, we can use a shortcut by looking at the leading and constant terms:

• The leading term \( z^3 \) divided by \( z^2 \) leaves \( z \).
• The constant term \( 106 \) divided by \( 53 \) leaves \( +2 \).

Therefore, the linear factor is \( (z + 2) \). This means the cubic equation fully factored is:

\[ (z^2 + 4z + 53)(z + 2) = 0 \]

Setting the linear factor to zero gives us the real root:

\[ z + 2 = 0 \implies \mathbf{z = -2} \]

a) To show that \( 2i \) is a root, we substitute \( z = 2i \) into the polynomial and verify that it equals zero.

First, calculate the powers of \( 2i \):

• \( (2i)^2 = 4i^2 = -4 \)
• \( (2i)^3 = 8i^3 = 8(-i) = -8i \)
• \( (2i)^4 = 16i^4 = 16(1) = 16 \)

Substitute these into the polynomial:

\[ (2i)^4 + (2i)^3 + 2(2i)^2 + 4(2i) - 8 \]

\[ = 16 + (-8i) + 2(-4) + 8i - 8 \]

\[ = 16 - 8i - 8 + 8i - 8 \]

\[ = (16 - 8 - 8) + (-8i + 8i) = \mathbf{0} \]

Since the result is 0, \( 2i \) is indeed a root.

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b) Because the polynomial has real coefficients, the Complex Conjugate Root Theorem states that \( -2i \) is also a root.

The factors corresponding to these two roots are \( (z - 2i) \) and \( (z + 2i) \). Multiplying them together gives a quadratic factor:

\[ (z - 2i)(z + 2i) = z^2 - (2i)^2 = z^2 - 4(-1) = z^2 + 4 \]

To find the remaining roots, we divide the original degree 4 polynomial by this quadratic factor \( z^2 + 4 \). You can use polynomial long division:

\[ \dfrac{z^4 + z^3 + 2z^2 + 4z - 8}{z^2 + 4} = z^2 + z - 2 \]

So, the polynomial factors as:

\[ (z^2 + 4)(z^2 + z - 2) = 0 \]

The remaining roots come from solving the quadratic equation \( z^2 + z - 2 = 0 \). Factoring it:

\[ (z + 2)(z - 1) = 0 \]

This yields \( z = -2 \) and \( z = 1 \). Therefore, the complete set of roots for the equation is:

\[ \mathbf{z = 2i, \quad z = -2i, \quad z = 1, \quad z = -2} \]

Because the polynomial has strictly real coefficients, any complex roots must occur in conjugate pairs. Since we are given two complex roots, we immediately know the other two roots:

Given roots: \( 2 - i \) and \( 1 - i \)

Conjugate roots: \( 2 + i \) and \( 1 + i \)

We can construct the polynomial by multiplying the factors associated with these four roots:

\[ P(z) = (z - (2 - i))(z - (2 + i)) \cdot (z - (1 - i))(z - (1 + i)) \]

Let's multiply the conjugate pairs together first. This will result in two real-coefficient quadratics. Regrouping to use the difference of squares:

First pair:

\[ ((z - 2) + i)((z - 2) - i) = (z - 2)^2 - i^2 = (z^2 - 4z + 4) + 1 = \mathbf{z^2 - 4z + 5} \]

Second pair:

\[ ((z - 1) + i)((z - 1) - i) = (z - 1)^2 - i^2 = (z^2 - 2z + 1) + 1 = \mathbf{z^2 - 2z + 2} \]

Now, multiply these two quadratic polynomials together to get \( P(z) \):

\[ P(z) = (z^2 - 4z + 5)(z^2 - 2z + 2) \]

Distribute each term in the first polynomial across the second:

\[ = z^2(z^2 - 2z + 2) - 4z(z^2 - 2z + 2) + 5(z^2 - 2z + 2) \]

\[ = (z^4 - 2z^3 + 2z^2) - (4z^3 - 8z^2 + 8z) + (5z^2 - 10z + 10) \]

Combine like terms:

• \( z^4 \) term: \( z^4 \)
• \( z^3 \) terms: \( -2z^3 - 4z^3 = -6z^3 \)
• \( z^2 \) terms: \( 2z^2 + 8z^2 + 5z^2 = 15z^2 \)
• \( z \) terms: \( -8z - 10z = -18z \)
• Constant term: \( 10 \)

\[ P(z) = z^4 - 6z^3 + 15z^2 - 18z + 10 \]

By comparing this to the given format \( z^4 + a z^3 + b z^2 + c z + d \), we find our coefficients:

\[ \mathbf{a = -6, \quad b = 15, \quad c = -18, \quad d = 10} \]

Expanding a binomial to the power of 6 is tedious. Instead, we can convert the complex number to its polar form \( z = r(\cos\theta + i\sin\theta) \) and use De Moivre's Theorem: \( z^n = r^n (\cos(n\theta) + i\sin(n\theta)) \).

First, find the modulus \( r \) for the complex number \( z = \sqrt{3} + i \):

\[ r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

Next, find the argument (angle) \( \theta \):

\[ \tan\theta = \dfrac{y}{x} = \dfrac{1}{\sqrt{3}} \]

Since both components are positive, the angle is in the first quadrant: \( \theta = \dfrac{\pi}{6} \) (or 30°).

So, the polar form is \( z = 2\left(\cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6}\right) \).

Now, apply De Moivre's Theorem for \( n = 6 \):

\[ z^6 = 2^6 \left( \cos\left(6 \cdot \dfrac{\pi}{6}\right) + i\sin\left(6 \cdot \dfrac{\pi}{6}\right) \right) \]

\[ z^6 = 64 (\cos(\pi) + i\sin(\pi)) \]

We know that \( \cos(\pi) = -1 \) and \( \sin(\pi) = 0 \):

\[ z^6 = 64 (-1 + 0i) = -64 \]

The standard form is: \( -64 + 0i \).

Let the complex variable be \( z = x + yi \), where \( x \) and \( y \) represent Cartesian coordinates.

Substitute this into the given modulus equation:

\[ |(x + yi) - 2| = |(x + yi) + 2i| \]

Group the real and imaginary components inside the moduli:

\[ |(x - 2) + yi| = |x + (y + 2)i| \]

The modulus of a complex number \( a+bi \) is \( \sqrt{a^2+b^2} \). Apply this to both sides:

\[ \sqrt{(x - 2)^2 + y^2} = \sqrt{x^2 + (y + 2)^2} \]

Square both sides to remove the square roots:

\[ (x - 2)^2 + y^2 = x^2 + (y + 2)^2 \]

Expand the perfect square binomials on both sides:

\[ x^2 - 4x + 4 + y^2 = x^2 + y^2 + 4y + 4 \]

Subtract \( x^2 \), \( y^2 \), and \( 4 \) from both sides. They cancel out nicely, leaving:

\[ -4x = 4y \]

Divide by 4 to get the final Cartesian equation:

\( y = -x \)

Geometric Description: The locus is a straight line passing through the origin with a slope of -1. (Conceptually, \( |z - A| = |z - B| \) always describes the perpendicular bisector of the line segment connecting points A and B in the complex plane. Here, A is (2, 0) and B is (0, -2). Their perpendicular bisector is indeed the line y = -x.)

To find the cube roots, we first write the number \( 8i \) in polar form.

The modulus is \( 8 \), and since it lies on the positive imaginary axis, its angle is \( \frac{\pi}{2} \). The general polar form is:

\[ 8i = 8 \left(\cos\left(\dfrac{\pi}{2} + 2k\pi\right) + i\sin\left(\dfrac{\pi}{2} + 2k\pi\right)\right) \]

To find the cube roots (\( z = (8i)^{1/3} \)), we use De Moivre's root formula: take the cube root of the modulus, and divide the argument by 3:

\[ z_k = \sqrt[3]{8} \left( \cos\left(\dfrac{\pi/2 + 2k\pi}{3}\right) + i\sin\left(\dfrac{\pi/2 + 2k\pi}{3}\right) \right) \]

\[ z_k = 2 \left( \cos\left(\dfrac{\pi}{6} + \dfrac{2k\pi}{3}\right) + i\sin\left(\dfrac{\pi}{6} + \dfrac{2k\pi}{3}\right) \right) \]

We find the three distinct roots by substituting \( k = 0, 1, 2 \):

For k = 0:

\[ z_0 = 2 \left( \cos\dfrac{\pi}{6} + i\sin\dfrac{\pi}{6} \right) = 2 \left( \dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i \right) = \mathbf{\sqrt{3} + i} \]

For k = 1:

\[ \text{Angle} = \dfrac{\pi}{6} + \dfrac{2\pi}{3} = \dfrac{\pi}{6} + \dfrac{4\pi}{6} = \dfrac{5\pi}{6} \]

\[ z_1 = 2 \left( \cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6} \right) = 2 \left( -\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i \right) = \mathbf{-\sqrt{3} + i} \]

For k = 2:

\[ \text{Angle} = \dfrac{\pi}{6} + \dfrac{4\pi}{3} = \dfrac{\pi}{6} + \dfrac{8\pi}{6} = \dfrac{9\pi}{6} = \dfrac{3\pi}{2} \]

\[ z_2 = 2 \left( \cos\dfrac{3\pi}{2} + i\sin\dfrac{3\pi}{2} \right) = 2(0 + (-1)i) = \mathbf{-2i} \]

The three roots are: \( \sqrt{3} + i \), \( -\sqrt{3} + i \), and \( -2i \).