Question

Express the following expression in the form of a + ib
\[\dfrac{\left( 3+i\sqrt{5} \right)\left( 3-i\sqrt{5} \right)}{\left( \sqrt{3}+\sqrt{2}i \right)-\left( \sqrt{3}-i\sqrt{2} \right)}\]

Answer 1

Hint: First simplify it and also use identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ ,after that multiply it by its conjugate where conjugate means that if complex number is c + id then its conjugate is c – id.

Complete Step-by-Step solution:
In the question a fraction of complex number is given which is \[\dfrac{\left( 3+i\sqrt{5} \right)\left( 3-i\sqrt{5} \right)}{\left( \sqrt{3}+\sqrt{2}i \right)-\left( \sqrt{3}-i\sqrt{2} \right)}\] and we have to express in for a + ib.
Before doing, so we will learn what complex numbers are.
A complex number is a number that can be written in form of a + bi, where a, b are real number and i is solution of equation ${{x}^{2}}=-1$ .This is because no real value satisfies for equation ${{x}^{2}}+1=0$ or ${{x}^{2}}=-1$ hence i is called imaginary number. For the complex number a + ib, a is considered as the real part and b is the imaginary part. Despite the historical nomenclature “imaginary, complex numbers are regarded in the mathematical sciences as just as “real” as real numbers, and are fundamental in any aspect of scientific description of the natural world.
Now it was given as \[\dfrac{\left( 3+i\sqrt{5} \right)\left( 3-i\sqrt{5} \right)}{\left( \sqrt{3}+\sqrt{2}i \right)-\left( \sqrt{3}-i\sqrt{2} \right)}\] by apply the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ and doing suitable calculation we get,
\[\dfrac{9-5{{i}^{2}}}{\sqrt{3}+\sqrt{2}i-\sqrt{3}+\sqrt{2}i}\]
By further simplifying and using fact that \[{{i}^{2}}=-1\] we get,
$\dfrac{9+5}{2\sqrt{2}i}=\dfrac{14}{2\sqrt{2}i}=\dfrac{7}{\sqrt{2}i}$
Now we will multiply by $\left( -\sqrt{2}i \right)$ to both the numerator and denominator we get,
$\dfrac{7}{\sqrt{2}i}\times \dfrac{-\sqrt{2}i}{-\sqrt{2}i}=\dfrac{-7\sqrt{2}i}{-2{{i}^{2}}}$
Once again using fact that ${{i}^{2}}=-1$ we get,
$\dfrac{-7\sqrt{2}i}{2}$
$\dfrac{-7\sqrt{2}i}{2}$ can also be written as $0+\left( \dfrac{-7\sqrt{2}}{2} \right)i$
So, in a + bi form a = 0 and b = $-\dfrac{7\sqrt{2}}{2}$
 Hence, the value is $0+\left( \dfrac{-7\sqrt{2}}{2} \right)i$

Note: Students should always try to eliminate the terms with or containing ‘i’ in denominator rationalizing with its conjugate multiplying both numerator and denominator.