If z1 is a complex number other than -1 such that z 1-1 and z 2- z 1-1 z 1-1- then show that the real parts of z 2 is zero-

Let nbsp;z=x+iy. Then, z2=z1 1z1+1 #160; #160; #160;=x+iy 1x+iy+1 #160; #160; #160;=x 1+iyx+1+iy #215;x+1 iyx+1 iy #160; #160; #160;=x2+x ...

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Standard XII

Mathematics

Differentiation of a Determinant

If z1 is a co...

Question

If z₁ is a complex number other than −1 such that $|z_{1}| = 1$ and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real parts of z₂ is zero.

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Solution

Let $z = x + i y$ .
Then,
$z_{2} = \frac{z_{1} - 1}{z_{1} + 1} = \frac{x + i y - 1}{x + i y + 1} = \frac{(x - 1) + i y}{(x + 1) + i y} \times \frac{(x + 1) - i y}{(x + 1) - i y} = \frac{x^{2} + x - i x y - x - 1 + i y + i x y + i y - i^{2} y^{2}}{{(x + 1)}^{2} - i^{2} y^{2}} = \frac{x^{2} + y^{2} - 1 + 2 i y}{x^{2} + 1 + 2 x + y^{2}} [∵ i^{2} = - 1]$

Now,
$Re (z_{2}) = \frac{x^{2} + y^{2} - 1}{x^{2} + y^{2} + 1 + 2 x} = 0 [∵ |z_{1}| = 1 \Rightarrow x^{2} + y^{2} = 1]$

Thus, the real parts of z₂ is zero.

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If $| z_{1} |$ is a complex number other than -1 such that $| z_{1} | = 1 and z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real parts of $z_{2}$ is zero.

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If z1 is a complex number other than −1 such that z1=1 and z2=z1-1z1+1, then show that the real parts of z2 is zero.

Let z=x+iy. Then, z2=z1-1z1+1 =x+iy-1x+iy+1 =x-1+iyx+1+iy×x+1-iyx+1-iy =x2+x-ixy-x-1+iy+ixy+iy-i2y2x+12-i2y2 =x2+y2-1+2iyx2+1+2x+y2 [∵ i2=-1] Now, Rez2=x2+y2-1x2+y2+1+2x =0 [∵ z1=1⇒x2+y2=1] Thus, the real parts of z2 is zero.

If z₁ is a complex number other than −1 such that $|z_{1}| = 1$ and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ , then show that the real parts of z₂ is zero.