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Miscellaneous Exercise · Q1

Q.Evaluate: [i18+(1i)25]3\left[i^{18} + \left(\dfrac{1}{i}\right)^{25}\right]^{3}.

Rajasthan RbseTextbookSubjective· 3mImportance★★★★★est
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The expression simplifies using the cyclic pattern of powers of ii and the fact that 1/i=−i1/i = -i. The final result is 2−2i\boxed{2 - 2i}.

Concept and Intuition

This problem is about complex number arithmetic — specifically, handling powers of the imaginary unit ii, where i=−1i = \sqrt{-1}. The key idea is that ii cycles through just four values: i,−1,−i,1i, -1, -i, 1 as its exponent increases by 1 each time. So any large exponent can be reduced by finding its remainder when divided by 4.

The second term, (1i)25\left(\frac{1}{i}\right)^{25}, might look tricky, but remember: 1i=−i\frac{1}{i} = -i (because i×(−i)=−i2=1i \times (-i) = -i^2 = 1). That turns it into a simple power of −i-i.

Once both terms inside the bracket are simplified, we cube the result. The cube of a complex number is just multiplication — but watch out for signs and the pattern of powers again.

Watch out

A common mistake is to forget that 1i=−i\frac{1}{i} = -i, not ii. Another is to misapply the exponent: (1i)25=1i25\left(\frac{1}{i}\right)^{25} = \frac{1}{i^{25}}, which is correct but then you must simplify i25i^{25} first.

Step-by-Step Solution

1. Simplify i18i^{18}.

The powers of ii repeat every 4:

i1=i,i2=−1,i3=−i,i4=1.i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1.

Divide 18 by 4: 18=4×4+218 = 4 \times 4 + 2, so remainder is 2. Hence:

i18=i2=−1.i^{18} = i^2 = -1.

2. Simplify (1i)25\left(\frac{1}{i}\right)^{25}.

First, note that 1i=−i\frac{1}{i} = -i (multiply numerator and denominator by ii: 1i⋅ii=i−1=−i\frac{1}{i} \cdot \frac{i}{i} = \frac{i}{-1} = -i). So:

(1i)25=(−i)25.\left(\frac{1}{i}\right)^{25} = (-i)^{25}.

Now (−i)25=(−1)25⋅i25=−1⋅i25(-i)^{25} = (-1)^{25} \cdot i^{25} = -1 \cdot i^{25} because (−1)25=−1(-1)^{25} = -1.

Reduce i25i^{25}: 25÷425 \div 4 gives remainder 1 (since 4×6=244 \times 6 = 24), so i25=i1=ii^{25} = i^1 = i. Thus:

(−i)25=−i.(-i)^{25} = -i.

Tip

Alternatively, use (1i)25=1i25\left(\frac{1}{i}\right)^{25} = \frac{1}{i^{25}}. Since i25=ii^{25} = i, we get 1i=−i\frac{1}{i} = -i directly. Same result, fewer steps.

3. Combine inside the bracket.

We have:

i18+(1i)25=(−1)+(−i)=−1−i.i^{18} + \left(\frac{1}{i}\right)^{25} = (-1) + (-i) = -1 - i.

4. Cube the result.

We need (−1−i)3(-1 - i)^3. Compute step by step:

(−1−i)2=(−1)2+2(−1)(−i)+(−i)2=1+2i+i2=2i.(-1 - i)^2 = (-1)^2 + 2(-1)(-i) + (-i)^2 = 1 + 2i + i^2 = 2i.

Now multiply by (−1−i)(-1 - i) again:

(−1−i)3=(2i)(−1−i)=−2i−2i2=2−2i.(-1 - i)^3 = (2i)(-1 - i) = -2i - 2i^2 = 2 - 2i.

Important

As a check, (−1−i)=−(1+i)(-1 - i) = -(1+i), so (−1−i)3=−(1+i)3(-1 - i)^3 = -(1+i)^3. Since (1+i)2=2i(1+i)^2 = 2i and (1+i)3=2i(1+i)=−2+2i(1+i)^3 = 2i(1+i) = -2 + 2i, we get −(1+i)3=2−2i-(1+i)^3 = 2 - 2i, confirming the result.

5. Final simplification (optional).

Factor out 2: 2−2i=2(1−i)2 - 2i = 2(1 - i). This is a perfectly acceptable final form.

✓Final answer

The value of the expression is 2−2i\boxed{2 - 2i} (or equivalently 2(1−i)\boxed{2(1 - i)}).

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