Questions on complex numbers are presented along with answers. Each solution is hidden inside a collapsible dropdown so you can practice independently before reviewing the detailed step-by-step explanations.
Add and express in the form of a complex number \( a + bi \) the expression:
\[ \dfrac{(2 + 3i) + (-4 + 5i) - (9 - 3i)}{3} \]Combine the real parts and the imaginary parts in the numerator:
Substitute back into the expression and divide by \( 3 \):
\[ \dfrac{-11 + 11i}{3} = -\dfrac{11}{3} + \dfrac{11}{3}i \]Multiply and express in the form of a complex number \( a + bi \) the expression:
\[ (-5 + 3i)(-4 + 8i) \]Expand using the distributive law (FOIL method):
\[ (-5)(-4) + (-5)(8i) + (3i)(-4) + (3i)(8i) \] \[ = 20 - 40i - 12i + 24i^2 \]Substitute \( i^2 = -1 \) and combine like terms:
\[ = 20 - 52i + 24(-1) = 20 - 24 - 52i = -4 - 52i \]Divide and express in the form of a complex number \( a + bi \) the expression:
\[ \dfrac{-1 - 2i}{-4 + 3i} \]Multiply the numerator and denominator by the complex conjugate of the denominator (\(-4 - 3i\)):
\[ \dfrac{(-1 - 2i)(-4 - 3i)}{(-4 + 3i)(-4 - 3i)} \]Expand the denominator: \((-4)^2 - (3i)^2 = 16 - 9(-1) = 25\).
Expand the numerator:
\[ (-1)(-4) + (-1)(-3i) + (-2i)(-4) + (-2i)(-3i) = 4 + 3i + 8i + 6i^2 = 4 + 11i + 6(-1) = -2 + 11i \]Divide by the denominator:
\[ \dfrac{-2 + 11i}{25} = -\dfrac{2}{25} + \dfrac{11}{25}i \]Find the complex conjugate of:
\[ 1 + 8i \]The conjugate of a complex number \( a + bi \) is \( a - bi \). Changing the sign of the imaginary part gives:
\[ 1 - 8i \]Express in the form of a complex number \( a + bi \):
\[ \dfrac{(-5 - i)(-7 + 8i)}{2 - 4i} \]First, expand the numerator:
\[ (-5)(-7) + (-5)(8i) + (-i)(-7) + (-i)(8i) = 35 - 40i + 7i - 8i^2 = 35 - 33i - 8(-1) = 43 - 33i \]Now divide by \( 2 - 4i \) by multiplying numerator and denominator by the conjugate \( 2 + 4i \):
\[ \dfrac{(43 - 33i)(2 + 4i)}{(2 - 4i)(2 + 4i)} = \dfrac{86 + 172i - 66i - 132i^2}{2^2 - (4i)^2} \] \[ = \dfrac{86 + 132 + 106i}{4 + 16} = \dfrac{218 + 106i}{20} = \dfrac{109}{10} + \dfrac{53}{10}i \]Express in the form of a complex number \( a + bi \):
\[ -(7 - i)(-4 - 2i)(2 - i) \]Multiply the first two terms and distribute the negative sign:
\[ (7 - i)(-4 - 2i) = -28 - 14i + 4i + 2i^2 = -28 - 10i - 2 = -30 - 10i \]Negate this result:
\[ -(-30 - 10i) = 30 + 10i \]Multiply by the final factor \( (2 - i) \):
\[ (30 + 10i)(2 - i) = 60 - 30i + 20i - 10i^2 = 60 - 10i - 10(-1) = 70 - 10i \]Express in the form of a complex number \( a + bi \):
\[ \dfrac{i}{1 - i} \]Multiply numerator and denominator by the conjugate \( 1 + i \):
\[ \dfrac{i(1 + i)}{(1 - i)(1 + i)} = \dfrac{i + i^2}{1^2 - i^2} = \dfrac{i - 1}{1 - (-1)} = \dfrac{-1 + i}{2} = -\dfrac{1}{2} + \dfrac{1}{2}i \]Solve for real \( x \) and \( y \) the equation:
\[ 2y + ix = 4 + x - i \]Equate the real parts and the imaginary parts from both sides:
Substitute \( x = -1 \) into the real equation:
\[ 2y = 4 + (-1) \implies 2y = 3 \implies y = \dfrac{3}{2} \]Solution: \( x = -1, y = \dfrac{3}{2} \)
Find \( a \) and \( b \), where \( a \) and \( b \) are real numbers so that:
\[ a + ib = (2 - i)^2 \]Expand the right side:
\[ (2 - i)^2 = 2^2 - 2(2)(i) + i^2 = 4 - 4i - 1 = 3 - 4i \]Equating real and imaginary parts yields \( a = 3 \) and \( b = -4 \).
Find the complex conjugate of:
\[ -3i \]Rewriting \(-3i\) in standard form as \(0 - 3i\), its complex conjugate is obtained by changing the sign of the imaginary part:
\[ 0 + 3i = 3i \]