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Exercise 4.1 · Q2

Q.Express the following in the form a+iba + ib: i9+i19i^{9} + i^{19}

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✓ Free question

Powers of ii cycle every four terms; reduce the exponents modulo 4 to find i9=ii^9 = i and i19=−ii^{19} = -i, giving a sum of 00.

Why this works: the periodicity of powers of ii

The imaginary unit ii is defined by i2=−1i^2 = -1. When you raise ii to successive powers, something beautiful happens: the results repeat in a cycle of length four. Understanding this pattern turns what looks like a tedious calculation into a simple lookup.

The fundamental cycle is:

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

After i4=1i^4 = 1, the pattern repeats: i5=i⋅i4=i⋅1=ii^5 = i \cdot i^4 = i \cdot 1 = i, and so on. This means any power of ii can be reduced by finding the remainder when the exponent is divided by 4.

in=in mod 4i^n = i^{n \bmod 4}

Step-by-step solution

  1. Reduce i9i^9 using the cycle.

    Divide 9 by 4: 9=4⋅2+19 = 4 \cdot 2 + 1, so 9≡1(mod4)9 \equiv 1 \pmod{4}.

    Therefore i9=i1=ii^9 = i^1 = i.

  2. Reduce i19i^{19} using the cycle.

    Divide 19 by 4: 19=4⋅4+319 = 4 \cdot 4 + 3, so 19≡3(mod4)19 \equiv 3 \pmod{4}.

    Therefore i19=i3=−ii^{19} = i^3 = -i.

  3. Add the two results.

i9+i19=i+(−i)=0i^9 + i^{19} = i + (-i) = 0

  1. Express in standard form a+iba + ib.

    Since 0=0+0i0 = 0 + 0i, we have a=0a = 0 and b=0b = 0.

Tip

To quickly find n mod 4n \bmod 4, just look at the last two digits of nn and find the remainder. For instance, 19÷419 \div 4 leaves remainder 3, so i19=i3i^{19} = i^3.

✓Final answer

The expression in the form a+iba + ib is 0+0i\boxed{0 + 0i} or simply 0\boxed{0}.

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