Question

|Z1 divide by z2|=|z1divide by z2| provide|z2|is not equal 0 ​

Answer 1

Answer:

+21958

+5

CPhill's answer is correct and much shorter than mine.

However, I'm assuming that you have the property of |x/y| = |x|/|y| for real numbers and now you are to prove the similar case for complex numbers; that is, when z1 = a + bi and z2 = c + di,

you need to prove that |(a + bi)/(c + di)| = |a + bi| / |c + di|.

By definition: |x + yi| = √(x² + y²)

So (right side): |a + bi| / |c + di| = √[ (a² + b²) / √(c² + d²) ]

and, since those are all real numbers, it can be rewritten as √[ (a² + b²) / (c² + d²) ].

Now, the left side:

| (a + bi) / (c + di) |

First, we'll need to write this as one complex number in the form x + yi; so let's multiply the numerator and denominator by the conjugate of the denominator, c - di:

| (a + bi)(c - di) / ( (c + di)(c - di) ) |

= | (ac + bd + bci - bdi) / (c² + d²) |

Splitting these into the real part and the imaginary part:

= | (ac + bd)/(c² + d²) + (bc-ad)i/(c² + d²) |

This is now in the form x + yi where x = (ac + bd)/(c² + d²) and y = (bc - ad)/(c² + d²).

Thus, the above equals: √ [ ( (ac + bd)/(c² + d²) )² + ( (bc- ad)/(c² + d²) )² ]

And (ac + bd)² = a²c² + 2abcd + b²d² and (bc - ad)² = b²c² - 2abcd + a²d²

So, the numerator becomes a²c² + b²d² + b²c² + a²d² = a²c² + b²c² + b²d² + a²d²

= (factoring) (a² + b²)c² + (a² + b²)d² = (a² + b²)(c² + d²)

And the denominator is: (c² + d²)²

to give the fraction, still under a radical sign:

√ [ (a² + b²)(c² + d²) / (c² + d²)² ] = √ [ (a² + b²) / (c² + d²) ]

which is what we had for the right side.