If f z -7- z -1- z 2- wehre z -1-2 i- then - f z - isA- -z-2B- - - - C- 2- z -D- none of these

The correct option is A |z|/2 Let z=1+2i ⇒ |z|=√(1+4)=√(5) Now, f(z)=7 z/1 z^2=7 1 2i/1 (1+2i)^2 =6 2i/1 1 4i^2 4i=6 2i/4 4i =(3 i)(2+2i)/(2 2i)(2+2i) =6 2i ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

If f z =7- z ...

Question

$If f (z) = \frac{7 - z}{1 - z^{2}}, wehre z = 1 + 2 i, then | f (z) | is$

$\frac{| z |}{2}$

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$| z |$

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$2 | z |$

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none of these

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Solution

The correct option is A
$\frac{| z |}{2}$

$Let z = 1 + 2 i \Rightarrow | z | = \sqrt{1 + 4} = \sqrt{5} Now, f(z) = \frac{7 - z}{1 - z^{2}} = \frac{7 - 1 - 2 i}{1 - (1 + 2 i)^{2}} = \frac{6 - 2 i}{1 - 1 - 4 i^{2} - 4 i} = \frac{6 - 2 i}{4 - 4 i} = \frac{(3 - i) (2 + 2 i)}{(2 - 2 i) (2 + 2 i)} = \frac{6 - 2 i + 6 i - 2 i^{2}}{4 - 4 i^{2}} = \frac{6 + 4 i + 2}{4 + 4} = \frac{8 + 4 i}{8} = 1 + \frac{1}{2} i f (z) = 1 + \frac{1}{2} i ∴ | f (z) | = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{4 + 1}{4}} = \frac{\sqrt{5}}{2} = \frac{| z |}{2}$

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If f(z)=7−z1−z2,wehre z=1+2i,then |f(z)| is

The correct option is A |z|2 Let z=1+2i⇒ |z|=√1+4=√5Now, f(z)=7−z1−z2=7−1−2i1−(1+2i)2=6−2i1−1−4i2−4i=6−2i4−4i=(3−i)(2+2i)(2−2i)(2+2i)=6−2i+6i−2i24−4i2=6+4i+24+4=8+4i8=1+12if(z)=1+12i∴ |f(z)|=√1+14=√4+14=√52=|z|2

$If f (z) = \frac{7 - z}{1 - z^{2}}, wehre z = 1 + 2 i, then | f (z) | is$