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.

11 and 12 Grade Math 

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