Division of Complex Numbers

Division of complex numbers is not directly possible in the same way as with real numbers. For example, 1a+bi\dfrac{1}{a + bi} isn't considered a simplified or useful form because the imaginary unit ii appears in the denominator.

To resolve this, we convert the division into multiplication — by multiplying both numerator and denominator with the conjugate of the denominator. This clever trick removes ii from the denominator and gives us a real number instead.

Why does this work? Because multiplying a complex number by its conjugate always gives a real number — as you've seen in the previous proof.

Let's walk through simplifying 1a+bi\dfrac{1}{a + bi}:

=1a+biabiabi= \dfrac{1}{a + bi} \cdot \dfrac{a - bi}{a - bi}

This keeps the expression equal but transforms it.

=abi(a+bi)(abi)= \dfrac{a - bi}{(a + bi)(a - bi)}

=abia2+b2= \dfrac{a - bi}{a^2 + b^2}

So we've rewritten 1a+bi\dfrac{1}{a + bi} as:

=aa2+b2ba2+b2i= \dfrac{a}{a^2 + b^2} - \dfrac{b}{a^2 + b^2}i

And now it's in the standard complex form x+yix + yi, where both parts are real. This is how division of complex numbers is handled — by eliminating the imaginary part from the denominator using conjugates.

Example

Let's simplify 13+4i\dfrac{1}{3 + 4i}

=13+4i34i34i= \dfrac{1}{3 + 4i} \cdot \dfrac{3 - 4i}{3 - 4i}

=34i(3+4i)(34i)= \dfrac{3 - 4i}{(3 + 4i)(3 - 4i)}

=34i3216i2= \dfrac{3 - 4i}{3^2 - 16i^2}

=34i9+16= \dfrac{3 - 4i}{9 + 16}

=34i25= \dfrac{3 - 4i}{25}

=325425i= \dfrac{3}{25} - \dfrac{4}{25}i

Which is the standard form of writing a complex number.

Simplify the following

182i\dfrac{1}{-8 - 2i}

Simplify the following

113i\dfrac{1}{-13 - i}

Simplify the following

111+9i\dfrac{1}{-11 + 9i}

Simplify the following

7+8i12i\dfrac{7 + 8i}{-1 - 2i}

Simplify the following

3+14i414i\dfrac{-3 + 14i}{4 - 14i}


End of Lesson

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Conjugates of Complex Numbers
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Task 09 - Find Multiplicative Inverse of Complex Numbers