Key to "Is |a−iba+ib| equal to 1? If yes, how?"

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2022-01-17

Is $| \frac{a - i b}{a + i b} |$ equal to 1? If yes, how?

Answer & Explanation

nick1337

Expert2022-01-18Added 777 answers

Step 1
$| z | = | \bar{z} |$
since the vector $\bar{z}$ is a mirror copy of z , upon reflection into the real axis, which leaves its norm (length) unaltered.
The above assumes that $a, b \in R$ . In fact the same is not true is we have $a, b \in C$
We do have
$| z |^{2} = z \bar{z}$
and
$\bar{z_{1} + z_{2}} = \bar{z_{1}} + \bar{z_{2}} \bar{z_{1} z_{2}} = \bar{z_{1} z_{2}}$
But if we take $a = 1, b = i$ we find:
$| 1 - i \times i | = | 2 | = 2$
and
$| 1 + i \times i | = | 0 | = 0$

Vasquez

Expert2022-01-18Added 669 answers

Step 1

| \frac{a - i b}{a + b i} |

= \frac{| a - i b |}{| a + i b |}

= \frac{\sqrt{a^{2} + (- b)^{2}}}{\sqrt{a^{2} + b^{2}}}

= \underset{―}{1}

alenahelenash

Expert2022-01-24Added 556 answers

Step 1Yes, presuming a and b are real. Then this is the quotient of conjugates, which have the same magnitude, so the magnitude of the quotient is 1. The result is on the unit circle, at double the angle of the numerator.

| \frac{a - b i}{a + b i} | = \frac{| a - b i |}{| a + b i |} = \frac{\sqrt{a^{2} + b^{2}}}{\sqrt{a^{2} + b^{2}}} = 1

If a and b are non-real complex numbers then in general this magnitude won't be 1.

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