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Question: How do you simplify \(\dfrac{3-2i}{1-i}\) and write it in \(a+ib\) form?...

How do you simplify 3−2i1−i\dfrac{3-2i}{1-i} and write it in a+iba+ib form?

Explanation

Solution

To simplify the above complex number3−2i1−i\dfrac{3-2i}{1-i}, first of all, we are going to multiply and divide by 1+i1+i and then simplify the numerator and denominator of the above expression. Doing this will remove the complex number “iota” in the denominator and we can express the given complex number into a+iba+ib. The power of (iota) “i” that you require in solving the above problem is i2=−1{{i}^{2}}=-1.

Complete step-by-step solution:
The complex number given in the above problem that we are asked to simplify is as follows:
3−2i1−i\dfrac{3-2i}{1-i}
To simplify the above complex number, we are going to rationalize the above number. This process of multiplication and division by the conjugate of the denominator of a complex number is called rationalization. For rationalization, we are going to multiply and divide by (1+i)\left( 1+i \right).
⇒3−2i1−i×1+i1+i\Rightarrow \dfrac{3-2i}{1-i}\times \dfrac{1+i}{1+i}
In the denominator, we are multiplying (1+i)\left( 1+i \right) by (1−i)\left( 1-i \right) and also in the numerator, we are multiplying (1+i)\left( 1+i \right) by (3−2i)\left( 3-2i \right) we get,
⇒(3−2i)(1+i)(1−i)(1+i) =3+3i−2i−2i2(1−i)(1+i) =3+i−2i2(1−i)(1+i) \begin{aligned} & \Rightarrow \dfrac{\left( 3-2i \right)\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)} \\\ & =\dfrac{3+3i-2i-2{{i}^{2}}}{\left( 1-i \right)\left( 1+i \right)} \\\ & =\dfrac{3+i-2{{i}^{2}}}{\left( 1-i \right)\left( 1+i \right)} \\\ \end{aligned}
As you can see the denominator in the above complex number is of the form (a−b)(a+b)\left( a-b \right)\left( a+b \right) and we know there is an algebraic identity which is equal to:
(a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}
⇒3+i−2i212−i2\Rightarrow \dfrac{3+i-2{{i}^{2}}}{{{1}^{2}}-{{i}^{2}}}
In the above, we are going to use the value of i2=−1{{i}^{2}}=-1 in the above equation and we get,
⇒3+i−2(−1)12−(−1) =3+i+21+1 =5+i2 \begin{aligned} & \Rightarrow \dfrac{3+i-2\left( -1 \right)}{{{1}^{2}}-\left( -1 \right)} \\\ & =\dfrac{3+i+2}{1+1} \\\ & =\dfrac{5+i}{2} \\\ \end{aligned}
Writing the above complex number into a+iba+ib form we get,
⇒52+12i\Rightarrow \dfrac{5}{2}+\dfrac{1}{2}i
Hence, we have simplified the above complex number into 52+12i\dfrac{5}{2}+\dfrac{1}{2}i.

Note: A trick to solve this simplification of a complex number is to rationalize the denominator which has a complex term and that will reduce your expression to a+iba+ib form. Now, let us discuss what this rationalization is. In the rationalization, we will multiply and divide the conjugate of the denominator and then simplify. The conjugate of a+iba+ib is a−iba-ib. If the denominator is of the form a−iba-ib then rationalization is done by multiplying and dividing by a+iba+ib Also, whenever you see any complex number in the denominator either with positive or negative sign then rationalization is the method to simplify the complex number.