Explore: Identities with Complex Numbers Solutions

Identities apply to complex numbers just as they apply to real numbers.
Simplify each side of the equation and then compare like terms to determine the missing value.

Note

Problems that contain the imaginary unit are not polynomial identities because polynomials can only have real number coefficients.

Example 7

Determine the value of Q that makes the identity true.

$(4 - Q i) (3 + 2 i) = 26 - 13 i$

Implement

$12 + 8 i - 3 Q i - 2 Q i^{2} = 26 - 13 i 12 + 8 i - 3 Q i - 2 Q (- 1) = 26 - 13 i - 12 - 8 i - 12 - 8 i - 3 Q i + 2 Q = 14 - 21 i \begin{matrix} 2 Q = 14 & \frac{- 3 Q i}{- 3 i} = \frac{- 21 i}{- 3 i} \\ Q = 7 & Q = 7 ✓ \end{matrix}$

Explain

Distribute
Simplify
Compare terms

Example 8

Determine if the following expressions form an identity.

$\begin{matrix} 3 \sqrt{- 12} + 4 \sqrt{- 162} & and & 6 i (\sqrt{3} + 7 \sqrt{2}) - \sqrt{72} \end{matrix}$

$3 i \sqrt{12} + 4 i \sqrt{162} 3 i (2^{\frac{2}{2}} \cdot 3^{\frac{1}{2}}) + 4 i (2^{\frac{1}{2}} \cdot 9^{\frac{2}{2}}) 3 i (2 \sqrt{3}) + 4 i (9 \sqrt{2}) 6 i \sqrt{3} + 36 i \sqrt{2}$

$6 i \sqrt{3} + 42 i \sqrt{2} - \sqrt{72} 6 i \sqrt{3} + 42 i \sqrt{2} - (2^{\frac{1}{2}} \cdot 6^{\frac{2}{2}}) 6 i \sqrt{3} + 42 i \sqrt{2} - (6 \sqrt{2}) 6 i \sqrt{3} + 42 i \sqrt{2} - 6 \sqrt{2}$

This is not an identity because the expressions are not equal to one another.