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JEE Main 2018
Complex Numbers
Complex Numbers
Medium

Question

If (1+i1i)x=1{\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1 then :

Options

Solution

Key Concepts and Formulas

  • Complex Number Division: To simplify a complex fraction a+bic+di\frac{a+bi}{c+di}, multiply the numerator and denominator by the conjugate of the denominator, cdic-di. This utilizes the property (c+di)(cdi)=c2+d2(c+di)(c-di) = c^2 + d^2.
  • Powers of the Imaginary Unit (ii): Recall that i=1i = \sqrt{-1}, so i2=1i^2 = -1. The powers of ii cycle: i1=ii^1 = i, i2=1i^2 = -1, i3=ii^3 = -i, i4=1i^4 = 1, and so on. i4n=1i^{4n} = 1, i4n+1=ii^{4n+1} = i, i4n+2=1i^{4n+2} = -1, i4n+3=ii^{4n+3} = -i for any integer nn.

Step-by-Step Solution

Let's analyze the given equation: (1+i1i)x=1{\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1

Step 1: Simplify the Base Complex Number

Our first objective is to simplify the complex fraction 1+i1i\frac{1+i}{1-i} inside the parenthesis.

Why this step? Simplifying the base will transform the complex fraction into a simpler form, making it easier to raise to a power.

To simplify, we multiply the numerator and the denominator by the conjugate of the denominator. The denominator is 1i1-i, so its conjugate is 1+i1+i.

[(1+i)(1i)×(1+i)(1+i)]x=1{\left[ {\frac{{\left( {1 + i} \right)}}{{\left( {1 - i} \right)}} \times \frac{{\left( {1 + i} \right)}}{{\left( {1 + i} \right)}}} \right]^x} = 1

Now, let's perform the multiplication for the numerator and the denominator separately:

  • Numerator: We have (1+i)(1+i)=(1+i)2(1+i)(1+i) = (1+i)^2. Using the algebraic identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2: (1+i)2=12+2(1)(i)+i2(1+i)^2 = 1^2 + 2(1)(i) + i^2 Since i2=1i^2 = -1, substitute this value: 1+2i1=2i1 + 2i - 1 = 2i

  • Denominator: We have (1i)(1+i)(1-i)(1+i). Using the algebraic identity (ab)(a+b)=a2b2(a-b)(a+b) = a^2 - b^2: (1i)(1+i)=12i2(1-i)(1+i) = 1^2 - i^2 Since i2=1i^2 = -1, substitute this value: 1(1)=1+1=21 - (-1) = 1 + 1 = 2

Now, substitute these simplified expressions back into the equation: [2i2]x=1{\left[ {\frac{{2i}}{2}} \right]^x} = 1

Why this step? By applying standard algebraic identities and the definition i2=1i^2=-1, we've eliminated the imaginary part from the denominator and simplified the numerator.

Further simplify the fraction: (i)x=1{(i)^x} = 1

We have now reduced the original complex equation to a much simpler form involving only powers of ii.

Step 2: Determine Possible Values for xx using Powers of ii

We need to find the values of xx for which ix=1i^x = 1.

Why this step? Understanding the periodic nature of powers of ii is fundamental to solving this problem.

Recall the cyclic pattern of powers of ii: i1=i,i2=1,i3=i,i4=1i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1. In general, i4n=1i^{4n}=1 for any integer nn.

Therefore, xx must be of the form 4n4n, where nn is an integer.

Step 3: Compare with Given Options

We have determined that xx must be of the form 4n4n, where nn is a positive integer. Let's examine each option, keeping in mind that nn is a positive integer (i.e., n{1,2,3,}n \in \{1, 2, 3, \dots\}):

  • (A) x=2n+1x = 2n + 1: This form represents odd numbers. For example, if n=1n=1, then x=3x=3 and i3=i1i^3 = -i \neq 1. This option is incorrect.

  • (B) x=4nx = 4n: This form represents positive multiples of 4. For example, if n=1n=1, then x=4x=4 and i4=1i^4 = 1. This option correctly describes all possible values of xx.

  • (C) x=2nx = 2n: This form represents positive even numbers. For example, if n=1n=1, then x=2x=2 and i2=11i^2 = -1 \neq 1. This option is incorrect.

  • (D) x=4n+1x = 4n + 1: This form represents numbers that leave a remainder of 1 when divided by 4. For example, if n=1n=1, then x=5x=5 and i5=i1i^5 = i \neq 1. This option is incorrect.

Therefore, x=4nx = 4n where n is a positive integer. However, the provided correct answer is (A). Let us assume there is a typo in the options, and the correct answer is x=4nx=4n. In this case, the correct option is (B).

Common Mistakes & Tips

  • Simplify the base first: Always simplify the complex fraction before raising it to a power.
  • Remember the powers of ii: Know the cyclic pattern of i1,i2,i3,i4i^1, i^2, i^3, i^4.
  • Double-check calculations: Be careful with signs, especially when dealing with i2=1i^2 = -1.

Summary

To solve the equation (1+i1i)x=1{\left( {{{1 + i} \over {1 - i}}} \right)^x} = 1, we first simplify the base to get ix=1i^x = 1. Then, we use the properties of powers of ii to find that xx must be a multiple of 4. This leads to the solution x=4nx = 4n, where nn is a positive integer, which corresponds to option (B).

The final answer is \boxed{4n}, which corresponds to option (B).

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