Question Video: Solving Equations Which Require Simplifying Quotients of Complex Numbers | Nagwa Question Video: Solving Equations Which Require Simplifying Quotients of Complex Numbers | Nagwa

Question Video: Solving Equations Which Require Simplifying Quotients of Complex Numbers Mathematics • First Year of Secondary School

Solve the equation 𝑧(2 + 𝑖) = 3 − 𝑖 for 𝑧.

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Video Transcript

Solve the equation 𝑧 multiplied by two plus 𝑖 is equal to three minus 𝑖 for 𝑧.

We begin here by dividing both sides of our equation by two plus 𝑖. On the left-hand side, this cancels, leaving us with 𝑧 is equal to three minus 𝑖 divided by two plus 𝑖. When dividing any two complex numbers, we need to multiply the numerator and denominator by the complex conjugate of the denominator, as this results in a purely real number on the denominator. If the complex number 𝑧 is equal to 𝑥 plus 𝑦𝑖, then its conjugate 𝑧 bar is equal to 𝑥 minus 𝑦𝑖. The conjugate of two plus 𝑖 is therefore equal to two minus 𝑖.

Our next step is to multiply these two fractions by distributing the parentheses on the numerator and then separately on the denominator. One way of doing this is using the FOIL method. Multiplying the first terms on the numerator gives us six. The outer terms have a product of negative three 𝑖, the inner terms a product of negative two 𝑖, and multiplying the last terms gives us positive 𝑖 squared.

Repeating this on the denominator gives us four minus two 𝑖 plus two 𝑖 minus 𝑖 squared. Negative two 𝑖 plus two 𝑖 is equal to zero. We also recall at this stage that 𝑖 squared is equal to negative one. The numerator therefore simplifies to five minus five 𝑖, and the denominator simplifies to five. All three of these terms are divisible by five. Therefore, the right-hand side simplifies to one minus 𝑖.

The complex number 𝑧 such that 𝑧 multiplied by two plus 𝑖 is equal to three minus 𝑖 is one minus 𝑖. We can check this answer by going back to our question and multiplying one minus 𝑖 by two plus 𝑖. Distributing our parentheses here gives us two plus 𝑖 minus two 𝑖 minus 𝑖 squared. Once again, as 𝑖 squared is equal to negative one, this simplifies to three minus 𝑖. This is indeed equal to the right-hand side of our original equation. Therefore, 𝑧 is definitely equal to one minus 𝑖.

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