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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