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Complex Numbers and Quadratic Equations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers and Quadratic Equations1 Mark
MCQ
If $\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2},$ where $\text{z}=1+2\text{i},$ then $|\text{f(z)}|$ is:
  • ✓
    $\frac{|\text{z}|}{2}$
  • B
    $|\text{z}|$
  • C
    $2|\text{z}|$
  • D
    None of these
✓

Answer

Correct option: A.
$\frac{|\text{z}|}{2}$
$\text{f(z)}=\frac{7-\text{z}}{1-\text{z}^2}$
$=\frac{7-(1+2\text{i})}{1-(1+2\text{i})^2}$
$=\frac{7-1+2\text{i}}{1-(1^2+2^2\text{i}^2+4\text{i})}$
$=\frac{6-2\text{i}}{1-1+4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}$
$=\frac{6-2\text{i}}{4-4\text{i}}\times\frac{4+4\text{i}}{4+4\text{i}}$
$=\frac{24+24\text{i}-8\text{i}-8\text{i}^2}{4^2-4^2\text{i}^2}$
$=\frac{24+16\text{i}+8}{16+16}$
$=\frac{32+16\text{i}}{32}$
$=1+\frac{1}{2}\text{i}$
since $\text{z}=1+2\text{i},$
$\because|\text{z}|=\sqrt{(1)^2+(2)^2}$
$=\sqrt{1+4}$
$=\sqrt{5}$
$\therefore|\text{f}\text{(z)|}=\sqrt{(1)^1+(\frac{1}{2})^2}$
$=\sqrt{1+\frac{1}{4}}$
$=\frac{\sqrt5}{2}$
$=\frac{|\text{z}|}{2}$

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