Question Video: Solving Equations with Complex Numbers | Nagwa Question Video: Solving Equations with Complex Numbers | Nagwa

Question Video: Solving Equations with Complex Numbers Mathematics • First Year of Secondary School

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

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Video Transcript

Solve the equation 𝑧 multiplied by two plus 𝑖 equals three minus 𝑖 for 𝑧.

To solve this equation for 𝑧, we’ll need to apply inverse operations. We’ll begin by dividing both sides of this equation by two plus 𝑖. And we see that 𝑧 is equal to three minus 𝑖 divided by two plus 𝑖. To divide three minus 𝑖 by two plus 𝑖, we’re going to need to multiply both the numerator and the denominator of the fraction by the conjugate of two plus 𝑖. To find the conjugate, we change the sign of the imaginary part. And we see that the conjugate of two plus 𝑖 is two minus 𝑖.

We’ll distribute the parentheses at the top of this fraction by using the FOIL method. Three multiplied by two is six. Three multiplied by negative 𝑖 is negative three 𝑖. We then get negative two 𝑖. And our last term gives us 𝑖 squared. 𝑖 squared is of course negative one. So our last term is negative one.

And we can collect like terms or add the real parts and separately add the imaginary parts. And we see that three minus 𝑖 multiplied by two minus 𝑖 is five minus five 𝑖. And we could repeat this process for the denominator.

However, there is a special rule we can use to multiply a complex number by its conjugate. We can find the sum of the squares of the real and imaginary parts. The real part is two, and the imaginary part, the coefficient of 𝑖, is one. So the product of these two complex numbers is four plus one, which is five. And we see that 𝑧 is equal to five minus five 𝑖 over five.

We can then divide the real parts by this real number. We get five divided by five, which is one. And separately we divide the imaginary part by this real number. Five divided by five is one. So we get one minus 𝑖. And we’ve solved our equation for 𝑧.

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