If $a$ and $b$ are real numbers such that $a^2+b^2=1$, then show that a real value of $x$ will satisfy the equation $\dfrac{1-ix}{1+ix} = (a-ib)$.

**OR**

If $z_1$ is a complex number other than $-1$ such that $|z_1| = 1$ and $z_2 = \dfrac{z_1-1}{z_1+1}$, then show that $z_2$ is

Question

If $a$ and $b$ are real numbers such that $a^2+b^2=1$, then show that a real value of $x$ will satisfy the equation $\dfrac{1-ix}{1+ix} = (a-ib)$.

**OR**

If $z_1$ is a complex number other than $-1$ such that $|z_1| = 1$ and $z_2 = \dfrac{z_1-1}{z_1+1}$, then show that $z_2$ is purely imaginary.

Accepted Answer

>

> [!TLDR]
> Cross-multiply, separate real and imaginary parts, and show the two resulting expressions for $x$ agree exactly because $a^2+b^2=1$.
>
> $$\frac{1-ix}{1+ix} = a-ib$$
> Cross-multiplying:
> $$1-ix = (a-ib)(1+ix) = a+iax-ib-i^2bx = a+iax-ib+bx = (a+bx)+i(ax-b)$$
>
> Equating real parts: $1 = a+bx \implies x = \dfrac{1-a}{b}$ (assuming $b
e0$).
>
> Equating imaginary parts: $-x = ax-b \implies -x(1+a) = -b \implies x = \dfrac{b}{1+a}$ (assuming $1+a
e0$).
>
> For $x$ to consistently satisfy both equations we need
> $$\frac{1-a}{b} = \frac{b}{1+a} \implies (1-a)(1+a) = b^2 \implies 1-a^2=b^2 \implies a^2+b^2=1$$
> which is exactly the given condition. So both expressions for $x$ agree, and since $a,b$ are real, $x=\dfrac{b}{1+a}$ is a real number satisfying the equation. Hence proved.
>
> ---
> **OR (alternative):** If $z_1
e-1$, $|z_1|=1$, $z_2=\dfrac{z_1-1}{z_1+1}$, show $z_2$ is purely imaginary.
>
> Let $z_1=x+iy$ with $x^2+y^2=1$.
> $$z_2 = \frac{(x-1)+iy}{(x+1)+iy}$$
> Multiply numerator and denominator by the conjugate of the denominator:
> $$z_2 = \frac{[(x-1)+iy][(x+1)-iy]}{(x+1)^2+y^2}$$ …

Q.If $a$ and $b$ are real numbers such that $a^2+b^2=1$ , then show that a real value of $x$ will satisfy the equation $\dfrac{1-ix}{1+ix} = (a-ib)$ . OR If $z_1$ is a complex number other than $-1$ such that $|z_1| = 1$ and $z_2 = \dfrac{z_1-1}{z_1+1}$ , then show that $z_2$ is purely imaginary.

Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

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