1.

What is the value of \(\rm [({i)^{25}+(\frac {1}{i})^{27}}]^2\), where i = \(\sqrt {-1}\)1. -22. \(\rm \frac 1 i\)3. -i4. -4

Answer» Correct Answer - Option 4 : -4

Concept:

Iota power:

  • i2 = -1
  • i3 = -i
  • i4 = 1

 

Law of exponents:

(am × an)  = am+n

\(\rm (a^m)^n=a^{mn}\)

 

Calculation:

Here, \(\rm [({i)^{25}+(\frac {1}{i})^{27}}]^2\)

= \(\rm [(i)^{25}+(\frac {1}{i})^{27}]^2\)                          ....(∵ √-1 = i)

= \(\rm [(i)^{24}i+(\frac {1}{i})^{24}(\frac 1 i)^3]^2\)                 .....(∵ (am × an)  = am+n)

= \(\rm [((i)^{6})^4i+((\frac {1}{i})^{6})^4(\frac 1 i)^3]^2\)           ....(∵ \(\rm (a^m)^n=a^{mn}\))

= \(\rm [i+(\frac 1 i)^3]^2\)                                 ....(∵ i4 = 1)

 \(=\rm [i-\frac 1 i]^2\\ =(\frac{i^2-1}{i})^2\)                                   ......(∵ i3 = -i)

\( \rm =(\frac{-1-1}{i})^2\\ =\frac{(-2)^2}{i^2}\)

= -4                                                ...(∵ i2 = -1)

Hence, option (4) is correct.


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