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Mathematics Question on Functions

IF f(z)=7−z1−z2f\left(z\right) = \frac{7-z}{1-z^{2}} , where z=1+2iz = 1 + 2i, then ∣f(z)∣|f(z)| is equal to :

A

∣z∣2\frac{|z|}{2}

B

∣z∣| z |

C

2∣z∣2| z |

D

None of these

Answer

∣z∣2\frac{|z|}{2}

Explanation

Solution

Given f(z)=7−z1−z2f(z)=\frac{7-z}{1-z^{2}}
where z=1+2iz=1+2 i
⇒∣z∣=12+22=5\Rightarrow|z|=\sqrt{1^{2}+2^{2}}=\sqrt{5}
⇒f(z)=y−z1−z2\Rightarrow f(z)=\frac{y-z}{1-z^{2}}
⇒y−(1+2i)1−(1+2i)2\Rightarrow \frac{y-(1+2 i)}{1-(1+2 i)^{2}}
=7−1−2i1−1−4i2−4i=\frac{7-1-2 i}{1-1-4 i^{2}-4 i}
⇒6−2i4−4i\Rightarrow \frac{6-2 i}{4-4 i}
=3−i2−2i=\frac{3-i}{2-2 i}
=3−i2−2i×2+2i2+2i=\frac{3-i}{2-2 i} \times \frac{2+2 i}{2+2 i}
=6+6i−2i−2i24−4i=\frac{6+6 i-2 i-2 i^{2}}{4-4 i}
=6+4i+24+4=\frac{6+4 i+2}{4+4}
⇒8+4i8\Rightarrow \frac{8+4 i}{8}
⇒1+1/2i\Rightarrow 1+1 / 2 i
⇒∣f(z)∣=12+(1/2)2\Rightarrow|f(z)|=\sqrt{1^{2}+(1 / 2)^{2}}
⇒1+1/4\Rightarrow \sqrt{1+1 / 4}
=52=\frac{\sqrt{5}}{2}
⇒∣z∣2\Rightarrow \frac{|z|}{2}