Problem 100 - Algebra and Trigonometry

Question

Complex Coefficients and the Quadratic Formula The quadratic formula works whether the coefficients of the equation are real or complex. Solve these equations using the quadratic formula and, if necessary, De Moivre's Theorem. $$ \begin{array}{l}{\text { (a) } z^{2}+(1+i) z+i=0} \\ {\text { (b) } z^{2}-i z+1=0} \\ {\text { (c) } z^{2}-(2-i) z-\frac{1}{4} i=0}\end{array} $$

Step-by-Step Solution

Answer

Use quadratic formula on each equation, applying De Moivre as needed. Compute discriminants to find complex roots.

Step 1: Identify coefficients for equation (a)

For the equation \( z^2 + (1+i)z + i = 0 \), identify the coefficients as follows: \( a = 1 \), \( b = 1+i \), and \( c = i \).

Step 2: Quadratic Formula for equation (a)

Apply the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = 1+i \), and \( c = i \) into the formula to find the roots of the equation.

Step 3: Calculate discriminant for equation (a)

The discriminant \( D = b^2 - 4ac \) must be calculated. Compute \( (1+i)^2 = 1 + 2i - 1 = 2i \) and \( 4ac = 4i \). Thus, \( D = 2i - 4i = -2i \).

Step 4: Solve equation (a) using discriminant

Substitute the values into the quadratic formula: \( z = \frac{-(1+i) \pm \sqrt{-2i}}{2} \). To handle \( \sqrt{-2i} \), apply De Moivre's Theorem by converting \(-2i\) to polar form and then finding the square roots. Finally, compute the two roots.

Step 5: Identify coefficients for equation (b)

For the equation \( z^2 - iz + 1 = 0 \), identify the coefficients as \( a = 1 \), \( b = -i \), and \( c = 1 \).

Step 6: Quadratic Formula for equation (b)

Use the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -i \), \( c = 1 \) to calculate the roots.

Step 7: Calculate discriminant for equation (b)

Compute the discriminant \( D = b^2 - 4ac = (-i)^2 - 4 \cdot 1 \cdot 1 = -1 - 4 = -5 \).

Step 8: Solve equation (b) using discriminant

Substitute into the formula to find \( z = \frac{i \pm \sqrt{-5}}{2} \). Use De Moivre’s Theorem to compute \( \sqrt{-5} \), and determine the two complex roots.

Step 9: Identify coefficients for equation (c)

For the equation \( z^2 - (2-i)z - \frac{1}{4}i = 0 \), the coefficients \( a = 1 \), \( b = -(2-i) \), \( c = -\frac{1}{4}i \) are identified.

Step 10: Quadratic Formula for equation (c)

Apply \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) by substituting the identified coefficients \( a = 1 \), \( b = -(2-i) \), \( c = -\frac{1}{4}i \).

Step 11: Calculate discriminant for equation (c)

Calculate \( D = b^2 - 4ac \) where \( b^2 = (2-i)^2 = 3 - 4i \), and \( 4ac = -i \). Thus, \( D = 3 - 4i + i = 3 - 3i \).

Step 12: Solve equation (c) using discriminant

Substitute \( D = 3 - 3i \) into the quadratic formula: \( z = \frac{2-i \pm \sqrt{3-3i}}{2} \). Convert \( 3-3i \) to polar form to find the square root. Compute and simplify to get the two roots.

Key Concepts

Complex Numbers

Complex numbers extend the real numbers by introducing an imaginary unit, denoted as \( i \), where \( i^2 = -1 \). A complex number is generally expressed in the form \( a + bi \), where \( a \) is the real part, and \( bi \) is the imaginary part. This allows us to perform arithmetic with numbers that include the square roots of negatives.
Applications of complex numbers are vast, especially in engineering, physics, and advanced mathematics.

• Addition of complex numbers involves adding the real parts together and the imaginary parts together. For example, \((2 + 3i) + (1 + 4i) = 3 + 7i \).
• Multiplication requires using the distributive property, ensuring \( i^2 = -1 \) is applied appropriately, such as \((1 + i)(1 - i) = 1 - i^2 = 1 + 1 = 2 \).
• Complex conjugates, expressed as \( \overline{a + bi} = a - bi \), help simplify division by removing imaginary units from the denominator when multiplying numerator and denominator by the conjugate.

Understanding complex numbers is essential when solving quadratic equations with non-real solutions.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in complex number calculations, especially for finding roots of complex numbers. This theorem states that for any complex number in polar form, \( z = r(\cos \theta + i \sin \theta) \), and any integer \( n \), \( z^n = r^n(\cos(n\theta) + i\sin(n\theta)) \).
This theorem simplifies handling powers and roots of complex numbers.

• To convert a complex number to polar form, calculate the magnitude \( r = \sqrt{a^2 + b^2} \) and the angle \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).
• For finding square roots, apply De Moivre's Theorem with \( n = \frac{1}{2} \), which will yield the principal root and its conjugate.
• This conversion and process are crucial when dealing with the square roots of complex numbers as discriminants in equations, like when calculating \( \sqrt{-2i} \) as seen in quadratic solutions.

Proficiency with De Moivre's Theorem enables the calculation of complex roots efficiently.

Discriminant

The discriminant is an important part of the quadratic formula. Given a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant is \( D = b^2 - 4ac \). It provides insight into the nature of the roots of the equation.

• If \( D > 0 \), the roots are distinct and real.
• If \( D = 0 \), the roots are real and identical.
• If \( D < 0 \), the roots are complex and conjugate pairs.

When the coefficients are complex, like in the equations \( z^2 + (1+i)z + i = 0 \) from our exercise, the discriminant becomes complex too.
To tackle the square roots of complex discriminants, De Moivre’s Theorem is usually applied. By converting the discriminant to polar form, one can find the roots easily even if they are not real.
The discriminant simplifies solving quadratic equations by allowing us to anticipate the nature of the roots, guiding the necessary steps in the solution.