This stack explains the process of dividing complex numbers. The key technique involves multiplying both the numerator and the denominator by the complex conjugate of the denominator to rationalize it, resulting in a standard 'a + bi' form.
What is the primary technique used to divide complex numbers, for example, \(\frac{a+bi}{c+di}\)?
Multiply both the numerator and the denominator by the complex conjugate of the denominator. This process is similar to rationalizing a denominator with a square root.
To simplify the expression \(\frac{3-i}{2+5i}\), what expression should you multiply the fraction by?
You should multiply by a fraction equal to \(1\), formed by the conjugate of the denominator over itself:$$ \frac{2-5i}{2-5i} $$
Why does multiplying by the conjugate of the denominator work when dividing complex numbers?
Multiplying a complex number by its conjugate, \((c+di)(c-di)\), always results in a real number \((c^2+d^2)\). This eliminates the imaginary unit \(i\) from the denominator.
Simplify the expression \(\frac{10}{3-i}\).
Multiply numerator and denominator by \(3+i\).
Numerator: \(10(3+i) = 30+10i\).
Denominator: \((3-i)(3+i) = 3^2+1^2 = 10\).
Result: \(\frac{30+10i}{10} = 3+i\).
Perform the division: \(\frac{2+8i}{4i}\).
The conjugate of \(4i\) is \(-4i\). Multiply by \(\frac{-4i}{-4i}\).
Numerator: \((2+8i)(-4i) = -8i - 32i^2 = 32-8i\).
Denominator: \((4i)(-4i) = -16i^2 = 16\).
Result: \(\frac{32-8i}{16} = 2 - \frac{1}{2}i\).
What is the multiplicative inverse (or reciprocal) of a complex number \(z = a+bi\)?
The multiplicative inverse is \(\frac{1}{z} = \frac{1}{a+bi}\). To simplify, multiply the numerator and denominator by the conjugate \(a-bi\), which gives \(\frac{a-bi}{a^2+b^2}\).
Find the multiplicative inverse of \(z = 4+3i\).
The inverse is \(\frac{1}{4+3i}\). Multiply by \(\frac{4-3i}{4-3i}\) to get \(\frac{4-3i}{4^2+3^2} = \frac{4-3i}{25}\). In standard form, this is \(\frac{4}{25} - \frac{3}{25}i\).
Simplify \(\frac{1}{i}\).
Multiply numerator and denominator by the conjugate of \(i\), which is \(-i\). This gives \(\frac{1 \cdot (-i)}{i \cdot (-i)} = \frac{-i}{-i^2} = \frac{-i}{1} = -i\).
