# Complex Number in the Standard Form

We will learn how to expand a complex in the standard form a + ib.

The following steps will help us to express a complex number in the standard form:

Step I: Obtain the complex number in the form $$\frac{a + ib}{c + id}$$ by using fundamental operations of addition, subtraction and multiplication.

Step II: Multiply the numerator and denominator by the conjugate of the denominator.

Solved examples on complex number in the standard form:

1. Express $$\frac{1}{2 - 3i}$$ in the standard form a + ib.

Solution:

We have $$\frac{1}{2 - 3i}$$

Now multiply the numerator and denominator by the conjugate of the denominator i.e., (2 + 3i), we get

= $$\frac{1}{2 - 3i}$$ × $$\frac{2 + 3i}{2 + 3i}$$

= $$\frac{2 + 3i}{2^{2} - 3^{2}i^{2}}$$

= $$\frac{2 + 3i}{4 + 9}$$

= $$\frac{2 + 3i}{13}$$

= $$\frac{2 }{13}$$ + $$\frac{3}{13}$$i, which is the required answer in a + ib form.

2. Express the complex number $$\frac{1 - i}{1 + i}$$ in the standard form a + ib.

Solution:

We have $$\frac{1 - i}{1 + i}$$

Now multiply the numerator and denominator by the conjugate of the denominator i.e., (1 - i), we get

= $$\frac{1 - i}{1 + i}$$ × $$\frac{1 - i}{1 - i}$$

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

= $$\frac{1 - 2i + i^{2}}{1 + 1}$$

= $$\frac{1 - 2i - 1}{2}$$

= $$\frac{- 2i }{2}$$

= - i

= 0 + (- i), which is the required answer in a + ib form.

3. Perform the indicated operation and find the result in the form a + ib.

$$\frac{3 - \sqrt{- 49}}{2 - \sqrt{-36}}$$

Solution:

$$\frac{3 - \sqrt{- 49}}{2 - \sqrt{-36}}$$

= $$\frac{3 - 7i}{2 - 6i}$$

Now multiply the numerator and denominator by the conjugate of the denominator i.e., (2 + 6i), we get

= $$\frac{3 - 7i}{2 - 6i}$$ × $$\frac{2 + 6i}{2 + 6i}$$

= $$\frac{(3 - 7i)(2 + 6i)}{2^{2} - 6^{2}i^{2}}$$

= $$\frac{6 + 18i - 14i - 42i^{2}}{4 + 36}$$

= $$\frac{6 + 4i + 42}{40}$$

= $$\frac{48 + 4i}{40}$$

= $$\frac{48 }{40}$$ + $$\frac{4}{40}$$i,

= $$\frac{6 }{5}$$ + $$\frac{1}{10}$$i, which is the required answer in a + ib form.

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