Explore a variety of complex number problems with step-by-step solutions. Learn how to solve complex numbers, including operations, polar form, and applications. Complex numbers play a crucial role in applied mathematics, physics, electrical engineering, and other technical fields.
In what follows, \( i \) is the imaginary unit.
Evaluate the following expressions:
If \( \dfrac{x + yi}{i} = 7 + 9i \), where \( x \) and \( y \) are real, what is the value of \( (x + yi)(x - yi) \)?
\[ \dfrac{x + yi}{i} = 7 + 9i \] \[ x + yi = i(7 + 9i) = -9 + 7i \] \[ (x + yi)(x - yi) = (-9 + 7i)(-9 - 7i) = 81 + 49 = 130 \]
Determine all complex numbers \( z \) that satisfy the equation: \[ z + 3z' = 5 - 6i \] where \( z' \) denotes the complex conjugate of \( z \).
Let \( z = a + bi \), and its conjugate \( z' = a - bi \); \( a \) and \( b \) real numbers
Substituting \( z \) and \( z' \) in the given equation obtain
\[ a + bi + 3(a - bi) = 5 - 6i \] \[ a + 3a + (b - 3b)i = 5 - 6i \] Simplify by grouping terms on theleft side: \[ 4 a - 2b i = 5 - 6i \] Two complex numbers are equal if both their real and imaginary parts are equal.. Hence \[ 4a = 5 \quad \text{and} \quad -2b = -6 \] \[ a = \dfrac{5}{4} \quad \text{and} \quad b = 3 \] \[ z = \dfrac{5}{4} + 3i \]Find all complex numbers of the form \( z = a + bi \), where \( a \) and \( b \) are real numbers, such that:
\[ z z' = 25 \quad \text{and} \quad a + b = 7 \] where, \( z' \) represents the complex conjugate of \( z \).Let \( z = a + b i \)
Hence its conjugate \[ z' = a - b i \] therefore \[ z z' = (a + b i)(a - b i) \] \[ = a^2 + b^2 = 25 \] \[ a + b = 7 \quad \text{gives} \quad b = 7 - a \] Substitute above in the equation \( a^2 + b^2 = 25 \) \[ a^2 + (7 - a)^2 = 25 \]Solve the above quadratic function for \( a \) and use \( b = 7 - a \) to find \( b \). \[ a = 4 \quad \text{and} \quad b = 3 \] or \[ a = 3 \quad \text{and} \quad b = 4 \] The complex numbers \[ z = 4 + 3i \] and \[ z = 3 + 4i \] satisfy \( z z' = 25 \).
Check that: \[ z = 4 + 3i \] and \[ z = 3 + 4i \] have the property \[ z z' = 25 \]
The complex number \( 2 + 4i \) is one of the roots of the quadratic equation
\[ x^2 + bx + c = 0, \] where \( b \) and \( c \) are real numbers.a) Find \( b \) and \( c \)
b) Write down the second root and check it.
a) Substitute the root in the equation: \[ (2 + 4i)^2 + b(2 + 4i) + c = 0 \] Expand terms in equation and rewrite as: \[ (-12 + 2b + c) + (16 + 4b)i = 0 \] Real part and imaginary part are both equal to zero. \[ -12 + 2b + c = 0 \quad \text{and} \quad 16 + 4b = 0 \] Solve for \( b \): \[ b = -4 \] Substitute and solve for \( c \): \[ c = 20 \]
b) Since the given equation has real numbers, the second root is the complex conjugate of the given root:
\( 2 - 4i \) is the second solution.
Check: \[ (2 - 4i)^2 - 4(2 - 4i) + 20 \] Expand: \[ = 4 - 16i + 16i - 16 + 8 - 20 + 20 \] \[ = (4 - 16 - 8 + 20) + (-16 + 16)i = 0 \]Find all complex numbers \( z \) such that: \[ z^2 = -1 + 2 \sqrt{6} i \]
Let \( z = a + bi \)
Substitute into given equation: \( (a + bi)^2 = -1 + 2\sqrt{6}i \)
Expand: \( a^2 - b^2 + 2abi = -1 + 2\sqrt{6}i \)
Real part and imaginary parts must be equal. \[ a^2 - b^2 = -1 \quad \text{and} \quad 2ab = 2\sqrt{6} \] Equation \( 2ab = 2\sqrt{6} \) gives: \( b = \dfrac{\sqrt{6}}{a} \)
Substitute: \( a^2 - \left( \dfrac{\sqrt{6}}{a} \right)^2 = -1 \) \[ a^4 - 6 = -a^2 \] Solve above equation and select only real roots: \( a = \sqrt{2} \) and \( a = -\sqrt{2} \)
Substitute to find \( b \) and write the two complex numbers that satisfy the given equation. \[ z_1 = \sqrt{2} + \sqrt{3}i, \quad z_2 = -\sqrt{2} - \sqrt{3}i \]
Find all complex numbers \( z \) such that \[ (4 + 2i)z + (8 - 2i)z' = -2 + 10i, \] where \( z' \) is the complex conjugate of \( z \).
Let \( z = a + bi \) where \( a \) and \( b \) are real numbers. The complex conjugate \( z' \) is written in terms of \( a \) and \( b \) as follows: \( z' = a - bi \).
Substitute \( z \) and \( z' \) in the given equation \[ (4 + 2i)(a + bi) + (8 - 2i)(a - bi) = -2 + 10i \] Expand and separate real and imaginary parts. \[ (4a - 2b + 8a - 2b) + (4b + 2a - 8b - 2a)i = -2 + 10i \] Two complex numbers are equal if their real parts and imaginary parts are equal. Group like terms. \[ 12a - 4b = -2 \quad \text{and} \quad -4b = 10 \] Solve the system of the unknown \( a \) and \( b \) to find: \[ b = -\dfrac{5}{2} \quad \text{and} \quad a = -1 \] \[ z = -1 - \dfrac{5}{2}i \]
Given that the complex number \( z = -2 + 7i \) is a root to the equation: \[ z^3 + 6z^2 + 61z + 106 = 0 \] find the real root to the equation.
Since \( z = -2 + 7i \) is a root of the equation and all the coefficients in the terms of the equation are real numbers, then \( z' \), the complex conjugate of \( z \), is also a solution. Hence we may factor the left side as follows: \[ z^3 + 6z^2 + 61z + 106 = (z - (-2 + 7i))(z - (-2 - 7i))q(z) \] \[ = (z^2 + 4z + 53)q(z) \] \[ q(z) = \dfrac{z^3 + 6z^2 + 61z + 106}{z^2 + 4z + 53} = z + 2 \] \( z + 2 \) is a factor of \( z^3 + 6z^2 + 61z + 106 \) and therefore \( z = -2 \) is the real root of the given equation.
a) Show that the complex number \( 2i \) is a root of the equation
\[ z^4 + z^3 + 2z^2 + 4z - 8 = 0 \] b) Find all the roots of this equation.a) Substitute \( z \) by \( 2i \) in the left side of the expression:
\[ z^4 + z^3 + 2z^2 + 4z - 8 \] \[ (2i)^4 + (2i)^3 + 2(2i)^2 + 4(2i) - 8 \] \[ = 16 - 8i - 8 + 8i - 8 = 0 \]which shows that \( 2i \) is a root of the given equation.
b) Since \( 2i \) is a root and all coefficients are real, \( -2i \) is also a root (complex conjugate). Hence we may factor the left side of the given equation as follows: \[ z^4 + z^3 + 2z^2 + 4z - 8 = (z - 2i)(z + 2i)q(z) \] \[ = (z^2 + 4)q(z) \] \[ q(z) = \dfrac{z^4 + z^3 + 2z^2 + 4z - 8}{z^2 + 4} = z^2 + z - 2 \]
The other two roots of the equation are the roots of \( q(z) = z^2 + z - 2\) and are given by: \[ z = 1 \quad \text{and} \quad z = -2 \].
\( P(z) = z^4 + a z^3 + b z^2 + c z + d \) is a polynomial where \( a \), \( b \), \( c \), and \( d \) are real numbers.
Find \( a \), \( b \), \( c \), and \( d \) if two zeros of polynomial \( P \) are the following complex numbers: \( 2 - i \) and \( 1 - i \).
Since all coefficients of polynomial P are real, the complex conjugate \[ 2 + i \quad \text{and} \quad 1 + i \] to the given zeros are also zeros of the polynomial \( P \). Hence \( P(z) \) in factored form: \[ P(z) = (z - (2 - i))(z - (2 + i))(z - (1 - i))(z - (1 + i)) = \] \[ = z^4 - 6 z^3 + 15 z^2 - 18 z + 10 \]
Identify \( a \), \( b \), \( c \), and \( d \) to obtain: \[ a = -6, \quad b = 15, \quad c = -18 , \quad d = 10. \]