The conjugate of a fancy quantity a + bi is the advanced quantity a – bi or a + -bi. The actual a part of the advanced quantity a + bi is a and the imaginary half is b.

a + bi and a – bi are known as advanced conjugates.

Discover that to seek out the advanced conjugate, all you should do is to take the other of the imaginary a part of the advanced quantity.

## Examples displaying how you can discover the advanced conjugate of a fancy quantity

**Instance #1**

Discover the advanced conjugate of 5 + 9i

9 is the imaginary half. The other of 9 is -9.

The advanced conjugate of 5 + 9i is 5 + -9i or 5 – 9i

**Instance #2**

Discover the advanced conjugate of -10 – 3i

-10 – 3i = -10 + -3i.

-3 is the imaginary half. The other of -3 is 3.

The advanced conjugate of -10 – 3i is -10 + 3i

**Instance #3**

Discover the advanced conjugate of i

i = 1i.

1 is the imaginary half. The other of 1 is -1.

The advanced conjugate of i is -i

The advanced conjugate of an actual quantity is the actual quantity since there isn’t any imaginary half. For instance, the advanced conjugate of two is 2.

## Why do we want the conjugate of a fancy quantity?

Allow us to see what’s going to occur once we multiply a fancy quantity by its advanced conjugate.

(a + bi)(a – bi) = a^{2} – abi + abi – b^{2}i^{2}

(a + bi)(a – bi) = a^{2 }– b^{2}i^{2}

(a + bi)(a – bi) = a^{2} – b^{2}(-1)

(a + bi)(a – bi) = a^{2} + b^{2}

Since i is just not right here anymore, a^{2} + b^{2} is only a actual quantity

**Instance**

(4 + 3i)(4 – 3i) = 4^{2} + 3^{2}

(4 + 3i)(4 – 3i) = 16 + 9

(4 + 3i)(4 – 3i) = 25

This property of the advanced conjugate will be very helpful if you find yourself attempting to simplify rational expressions or if you find yourself attempting to do away with the “i” within the denominator of a rational expression.

For instance, simplify 5i / (2 – i).

Multiply the numerator and the denominator of 5i / (2 – i) by the conjugate of two – i

The conjugate of two – i is 2 + i.

[5i(2 + i)] / (2 – i)(2 + i) = (10i + 5i^{2}) / (2^{2} + 1^{2})

[5i(2 + i)] / (2 – i)(2 + i) = [10i + 5(-1)] / ( 4 + 1)

[5i(2 + i)] / (2 – i)(2 + i) = (10i – 5) / 5

[5i(2 + i)] / (2 – i)(2 + i) = 2i – 1

Due to this fact, 5i / (2 – i) = 2i – 1