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The conjugate of a posh quantity a + bi is the complicated quantity a – bi or a + -bi. The true a part of the complicated quantity a + bi is a and the imaginary half is b.

a + bi and a – bi are referred to as complicated conjugates.

Discover that to search out the complicated conjugate, all you must do is to take the alternative of the imaginary a part of the complicated quantity.

## Examples displaying how one can discover the complicated conjugate of a posh quantity

**Instance #1**

Discover the complicated conjugate of 5 + 9i

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

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

**Instance #2**

Discover the complicated conjugate of -10 – 3i

-10 – 3i = -10 + -3i.

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

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

**Instance #3**

Discover the complicated conjugate of i

i = 1i.

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

The complicated conjugate of i is -i

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

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

Allow us to see what’s going to occur once we multiply a posh quantity by its complicated 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 just isn’t 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 complicated conjugate will be very helpful if you find yourself attempting to simplify rational expressions or if you find yourself attempting to eliminate 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

Subsequently, 5i / (2 – i) = 2i – 1

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