Question:

IF f(z)=7z1z2f\left(z\right) = \frac{7-z}{1-z^{2}} , where z=1+2iz = 1 + 2i, then f(z)|f(z)| is equal to :

Updated On: Jun 23, 2023
  • z2\frac{|z|}{2}
  • z| z |
  • 2z2| z |
  • None of these
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The Correct Option is A

Solution and Explanation

Given f(z)=7z1z2f(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)=yz1z2\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}}
=712i114i24i=\frac{7-1-2 i}{1-1-4 i^{2}-4 i}
62i44i\Rightarrow \frac{6-2 i}{4-4 i}
=3i22i=\frac{3-i}{2-2 i}
=3i22i×2+2i2+2i=\frac{3-i}{2-2 i} \times \frac{2+2 i}{2+2 i}
=6+6i2i2i244i=\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}
z2 \Rightarrow \frac{|z|}{2}
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Concepts Used:

Functions

A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.

Kinds of Functions

The different types of functions are - 

One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.

Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.

Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.

Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.

Read More: Relations and Functions