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COMPLEX NUMBERS AND QUADRATIC EQUATIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS1 Mark
MCQ
The least positive integer n such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer, is:
  • A
    16
  • ✓
    8
  • C
    4
  • D
    2
✓

Answer

Correct option: B.
8
Let $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)$
$\Rightarrow\text{z}=\frac{2\text{i}}{1+\text{i}}\times\frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1-\text{i}^2}$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{1+1} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{2\text{i}(1-\text{i})}{2}$
$\Rightarrow\text{z}=\text{i}-\text{i}^2$
$\Rightarrow\text{z}=\text{i}+1$
Now, $\text{z}^\text{n}=(1+\text{i})^\text{n}$
For $\text{n}=2,$
$\text{z}^2=(1+\text{i})^2$
$=1+\text{i}^2+2\text{i}$
$=1-1+2\text{i}$
$=2\text{i} \ ...(1)$
Since this is not a positive integer,
For $\text{n}=4,$
$\text{z}^4=(1+\text{i})^4$
$=\big[(1+\text{i})^2\big]^2$
$=(2\text{i})^2$ [Using (1)]
$=(4\text{i})^2$
$=-4 \ ...(2)$
This is a negative integer.
For $\text{n}=8,$
$\text{z}^8=(1+\text{i})^8$
$=\big[(1+\text{i})^4\big]^2$
$=(-4)^2$ [Using (2)]
$=16$
This is a positive integer.
Thus, $\text{z}=\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is positive for $\text{n}=8.$
Therefore, 8 is the least positive integer such that $\Big(\frac{2\text{i}}{1+\text{i}}\Big)^\text{n}$ is a positive integer.

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