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Worked Examples · Example 6

Q.Express the following in the form a+iba + ib:

(i) 5+2 i1−2 i\dfrac{5 + \sqrt{2}\,i}{1 - \sqrt{2}\,i}
(ii) i−35i^{-35}
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The key idea is to rationalise the denominator for part (i) by multiplying by the complex conjugate, and to use the cyclic pattern of powers of ii for part (ii). The results are 1+22 i1 + 2\sqrt{2}\,i and ii, respectively.

Concept and Intuition

Complex numbers are written as a+iba + ib, where aa and bb are real numbers. When you have a fraction with a complex denominator, the trick is to make the denominator real — because dividing by a real number is straightforward. This is done by multiplying numerator and denominator by the complex conjugate of the denominator. The conjugate of x+iyx + iy is x−iyx - iy; their product is x2+y2x^2 + y^2, a real number.

For powers of ii, remember that i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1, and then the pattern repeats every 4. So any large exponent can be reduced by looking at the remainder when divided by 4.

Step-by-step solution

Part (i): 5+2 i1−2 i\dfrac{5 + \sqrt{2}\,i}{1 - \sqrt{2}\,i}

1. Identify the conjugate of the denominator.

The denominator is 1−2 i1 - \sqrt{2}\,i. Its conjugate is 1+2 i1 + \sqrt{2}\,i.

2. Multiply numerator and denominator by this conjugate.

This does not change the value of the fraction, because we are multiplying by 11:

5+2 i1−2 i×1+2 i1+2 i\frac{5 + \sqrt{2}\,i}{1 - \sqrt{2}\,i} \times \frac{1 + \sqrt{2}\,i}{1 + \sqrt{2}\,i}

3. Simplify the denominator.

The denominator becomes a difference of squares:

(1−2 i)(1+2 i)=12−(2 i)2=1−(2⋅i2)(1 - \sqrt{2}\,i)(1 + \sqrt{2}\,i) = 1^2 - (\sqrt{2}\,i)^2 = 1 - (2 \cdot i^2)

Since i2=−1i^2 = -1, this is:

1−(2⋅(−1))=1+2=31 - (2 \cdot (-1)) = 1 + 2 = 3

Tip

Notice that multiplying a complex number by its conjugate always gives a2+b2a^2 + b^2 (a real number). Here a=1a=1, b=−2b=-\sqrt{2}, so a2+b2=1+2=3a^2+b^2 = 1 + 2 = 3. This saves you from expanding every time.

4. Simplify the numerator.

Expand (5+2 i)(1+2 i)(5 + \sqrt{2}\,i)(1 + \sqrt{2}\,i):

5⋅1+5⋅2 i+2 i⋅1+2 i⋅2 i5 \cdot 1 + 5 \cdot \sqrt{2}\,i + \sqrt{2}\,i \cdot 1 + \sqrt{2}\,i \cdot \sqrt{2}\,i

=5+52 i+2 i+2 i2= 5 + 5\sqrt{2}\,i + \sqrt{2}\,i + 2\,i^2

Now i2=−1i^2 = -1, so 2 i2=−22\,i^2 = -2. Combine the real parts: 5−2=35 - 2 = 3. Combine the imaginary parts: 52 i+2 i=62 i5\sqrt{2}\,i + \sqrt{2}\,i = 6\sqrt{2}\,i. So the numerator is:

3+62 i3 + 6\sqrt{2}\,i

5. Divide by the real denominator.

3+62 i3=1+22 i\frac{3 + 6\sqrt{2}\,i}{3} = 1 + 2\sqrt{2}\,i

Watch out

A common mistake is to forget that (2 i)2=2i2=−2(\sqrt{2}\,i)^2 = 2 i^2 = -2, not 22. Always handle i2i^2 carefully.

So part (i) gives 1+22 i1 + 2\sqrt{2}\,i.

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