Question

If $$a={\displaystyle \frac{z-1}{z+1}}$$ is purely imaginary number (z is not equal to$$-1$$), then mod $z$ is ?
A. $$1$$
B. $$2$$
C. $$3$$
D. $$4$$

Answer 1

Hint: A complex number is of the form a +ib where a and b are real numbers and $$i$$ is an imaginary number. Here, we will take $$z=x+iy$$ and substitute the value of z. We will also use the formula $$(a-b)(a+b)=a^2-b^2$$ and $$(a-b{)}^2=a^2+b^2-2ab$$. We know that $$i^2=-1$$. For the given complex number is in the form $$z=x+iy$$, then the conjugate of that number will be $$\bar{z}=x-iy$$. Here, if you take the conjugate number as a complex number and then solve it, you will still get the correct answer.

Complete step by step answer:
Let, $$z=x+iy$$ and $$z\ne -1$$.
Give that, $$a={\displaystyle \frac{z-1}{z+1}}$$
Substituting the value of z, we get,
$$a={\displaystyle \frac{x+iy-1}{x+iy+1}}$$
Rearrange the above expression, we get,
$$a={\displaystyle \frac{x-1+iy}{x+1+iy}}$$
Multiply to both the numerator and denominator with$$x+1-iy$$, we get,
$$\begin{gathered} a={\displaystyle \frac{x-1+iy}{x+1+iy}}\times {\displaystyle \frac{x+1-iy}{x+1-iy}} \\ \Rightarrow a={\displaystyle \frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}} \end{gathered}$$
$$\Rightarrow a={\displaystyle \frac{(x-(1-iy))(x+(1-iy))}{((x+1)+iy)((x+1)-iy)}}$$
Applying this formula$$(a-b)(a+b)=a^2-b^2$$, we get,
$$a={\displaystyle \frac{x^2-{(1-iy)}^2}{{(x+1)}^2-{(iy)}^2}}$$
Applying this formula$$(a-b{)}^2=a^2+b^2-2ab$$, we get,
Also, $$(iy{)}^2=i^2{y}^2=(-1)y^2=-y^2$$ $$(\because i^2=-1)$$
$$a={\displaystyle \frac{x^2-(1^2+{(iy)}^2-2(iy))}{{(x+1)}^2-{(iy)}^2}}$$
$$\Rightarrow a={\displaystyle \frac{x^2-(1-y^2-2yi)}{{(x+1)}^2-(-y^2)}}$$
$$\Rightarrow a={\displaystyle \frac{x^2-1+y^2+2yi}{{(x+1)}^2+y^2}}$$
$$\Rightarrow a={\displaystyle \frac{x^2+y^2-1}{{(x+1)}^2+y^2}}+{\displaystyle \frac{2y}{{(x+1)}^2+y^2}}i$$

Since it is given that $${\displaystyle \frac{z-1}{z+1}}$$ is purely imaginary, it means that the real part is zero.
$${\displaystyle \frac{x^2+y^2-1}{{(x+1)}^2+y^2}}=0$$
$$\begin{gathered} \Rightarrow x^2+y^2-1=0 \\ \Rightarrow x^2+y^2=1 \end{gathered}$$
We know that, $$\left|z\right|=\left|x+iy\right|=\sqrt{x^2+y^2}$$
$$\Rightarrow \sqrt{x^2+y^2}=\sqrt{1}$$
$$\Rightarrow \left|z\right|=\pm 1$$
Since, it is given that, $$z\ne -1$$
$$\therefore \left|z\right|=1$$

Note: Another method: Since it is given that$${\displaystyle \frac{z-1}{z+1}}$$is purely imaginary,
$${\displaystyle \frac{z-1}{z+1}}+(\overline{{\displaystyle \frac{z-1}{z+1}}})=0$$
$$\Rightarrow {\displaystyle \frac{z-1}{z+1}}+{\displaystyle \frac{\bar{z}-1}{\bar{z}+1}}=0$$
$$\Rightarrow {\displaystyle \frac{(z-1)(\bar{z}+1)+\overline{(z}-1)(z+1)}{(z+1)(\bar{z}+1)}}=0$$
$$\Rightarrow (z-1)(\bar{z}+1)+\overline{(z}-1)(z+1)=0$$
$$\begin{gathered} \Rightarrow z\bar{z}+z-\bar{z}-1+\bar{z}z+\bar{z}-z-1=0 \\ \Rightarrow 2z\bar{z}-2=0 \end{gathered}$$
$$\Rightarrow z\bar{z}-1=0$$
$$\Rightarrow {\left|z\right|}^2=1$$
$$\Rightarrow \left|z\right|=\pm 1$$
Since, it is given that, $$z\ne -1$$
$$\therefore \left|z\right|=1$$
Hence, if $${\displaystyle \frac{z-1}{z+1}}$$ is purely imaginary number and$$z\ne -1$$ then $$\left|z\right|=1$$.