Multiplying a complex number by its conjugate

In the same way that multiplying a bracket involving a surd by the conjugate of the bracket will produce a rational number, the same is try for multiplying a complex number by the complex conjugate.

Multiplying a complex number by its conjugate will always result in a real number.

Proof

Let z=x+yi. x and y are real.
z^*=x-yi
(x+yi)(x-yi)=x^2+xyi-xyi-y^2i^2
(x+yi)(x-yi)=x^2-y^2i^2
(x+yi)(x-yi)=x^2-y^2(-1)
(x+yi)(x-yi)=x^2+y^2
x and y are real, so x^2+y^2 is real.

Example: find z such that 3z+2z^*=5+2i

Let z=x+yi
z^*=x-yi
3(x+yi)+2(x-yi)=5+2i
3x+3yi+2x-2yi=5+2i
5x+yi=5+2i
5x=5, x=1
1y=2, y=2
z=x+yi
z=1+2i

flashcards

Question	Answer
What happens when you multiply a complex number by its conjugate?	It always results in a real number.
What is the proof that (x+yi)(x-yi) is real?	(x+yi)(x-yi) = x^2 - y^2i^2 = x^2 - y^2(-1) = x^2 + y^2, which is real because x and y are real.
How do you solve 3z + 2z^* = 5 + 2i for z?	Let z = x + yi, so z^* = x - yi. Then 3(x + yi) + 2(x - yi) = 5x + yi = 5 + 2i. Equate real and imaginary parts: 5x = 5 gives x = 1, and y = 2 gives z = 1 + 2i.
What is the conjugate of z = x + yi?	z^* = x - yi

Multiplying a complex number by its conjugate

Proof

Example: find z such that 3z+2z^*=5+2i

flashcards

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