Proving that a real value of 'x' will satisfy the equation

Question

Show that a real value of $x$ will satisfy the equation $\frac{1-ix}{1+ix}$ = $a-ib$ if $a² + b² = 1$

And it is given that $a, b$ are real.

Answer 1

Alternatively, note that ($ab\ne 0$): $$a² + b² = 1 \iff (a+bi)(a-bi)=1 \iff a-bi=\frac{1}{a+bi}=\frac{1-ix}{1+ix} \iff \\ 1+ix=(1-ix)(a+bi) \iff \\ 1+ix=a+bi-axi+bx \iff \\ x=\frac{a-1+bi}{-b+(a+1)i}=\frac{(a-1+bi)(-b-(a+1)i)}{b^2+(a+1)^2}=\\ \frac{\color{red}{-ab}-a^2i\color{blue}{-ai}+b+\color{blue}{ai}+i-b^2i+\color{red}{ab}+b}{b^2+(a+1)^2}=\\ \frac{-(a^2+b^2)i+i+2b}{b^2+(a+1)^2}=\\ \frac{2b}{b^2+(a+1)^2}.$$

Answer 2

We have an equation with $x$. Let's see what we can do:

$$1-ix = (1+ix)(a-bi)$$

$$1-ix = (a+bx) +i(ax-b)$$

Now we can try to assume that $x$ is real and at least get some candidates for $x$. With this assumption, we can now equate real and imaginary parts:

$1 = a+bx$

$-x= ax-b$

Now we have $2$ equations in one variable, so we better get the same solution from each of them.

And we do. We get $x = \dfrac {1-a}{b}$.

Now it remains to check that this $x$ satisfies the original equation. You may have to use that $a^2+b^2=1$.

Answer 3

Just multiply the denominator and numerator of the equation's RHS with $(1-ix)$, this would give you the values of $a$ and $b$

$$\frac {(1-ix)^2} {1-i^2x^2} = a -ib $$

$$\frac {1-x^2 - 2ix} {1+x^2} = a - ib$$

$$\frac {1-x^2} {1+x^2} - \frac {2ix} {1+x^2} = a- ib $$

Since it's given that a and b are real, you can compare like follows:

$$a = \frac {1-x^2} {1+x^2}$$ and $$\frac {2x} {1+x^2} = b$$

Now you have $a^2 + b^2 = 1$, solve this using the above values of $a$ and $b$

$$ (1-x^2)^2 + 4x^2 = (1+x^2)^2$$

You may solve this to find that real values of x satisfy your question.

Answer 4

$\cos(\theta)=\frac{1}{\sqrt{1+x^2}},\sin(\theta)=\frac{x}{\sqrt{1+x^2}},\tan(\theta)=x$

$a-bi=\frac{1-ix}{1+ix}=\frac{\frac{1-ix}{\sqrt{1+x^2}}}{\frac{1+ix}{\sqrt{1+x^2}}}=\exp(-2\theta i)=\exp(i\arctan(\frac{-b}{a}))$

$2\theta+2k\pi=\arctan(\frac{b}{a}),\tan(2\theta)=\frac{2x}{1-x^2}=\frac{b}{a}$