Solveeit Logo
Login

Question

Question: How do you express \({{\left( 1-i \right)}^{3}}\) in \(a+bi\) form?...

How do you express (1−i)3{{\left( 1-i \right)}^{3}} in a+bia+bi form?

Explanation

Solution

In this problem we have to express (1−i)3{{\left( 1-i \right)}^{3}} in the form of a+bia+bi. There are different methods to solve this problem. But for now, we will go with the method of factorization. In the problem we have(1−i)3{{\left( 1-i \right)}^{3}}. So, we will write it as (1−i)3=(1−i)(1−i)(1−i){{\left( 1-i \right)}^{3}}=\left( 1-i \right)\left( 1-i \right)\left( 1-i \right) according to exponential rule a.a.a.a.a.... n times=ana.a.a.a.a....\text{ n times}={{a}^{n}}. Now we will multiply the first two terms by using the distribution law of multiplication over subtraction and simplify the equation by substituting whenever we got i2{{i}^{2}} with −1-1 because we know that i2=−1{{i}^{2}}=-1. After simplifying the product of the two terms we will again multiply it with the third term to get the final result. In this step also we will follow the above procedure to simplify the equation.

Complete step by step solution:
Given that, (1−i)3{{\left( 1-i \right)}^{3}}.
Now we will expand the above expression by using the exponential rule a.a.a.a.a.... n times=ana.a.a.a.a....\text{ n times}={{a}^{n}}, then we will get
(1−i)3=(1−i)(1−i)(1−i){{\left( 1-i \right)}^{3}}=\left( 1-i \right)\left( 1-i \right)\left( 1-i \right)
Multiplying the two terms in the above equation, then we will get
⇒(1−i3)=(1−i−i+i2)(1−i)\Rightarrow \left( 1-{{i}^{3}} \right)=\left( 1-i-i+{{i}^{2}} \right)\left( 1-i \right)
Simplifying the above equation, then we will have
⇒(1−i3)=(1−2i+i2)(1−i)\Rightarrow \left( 1-{{i}^{3}} \right)=\left( 1-2i+{{i}^{2}} \right)\left( 1-i \right)
We know that i2=−1{{i}^{2}}=-1, Substituting this value in the above equation, then we will get
⇒(1−i3)=(1−2i+(−1)2)(1−i)\Rightarrow \left( 1-{{i}^{3}} \right)=\left( 1-2i+{{\left( -1 \right)}^{2}} \right)\left( 1-i \right)
We know that (−1)2=1{{\left( -1 \right)}^{2}}=1, then we will get
⇒(1−i3)=(1−2i−1)(1−i) ⇒(1−i3)=(−2i)(1−i) \begin{aligned} & \Rightarrow \left( 1-{{i}^{3}} \right)=\left( 1-2i-1 \right)\left( 1-i \right) \\\ & \Rightarrow \left( 1-{{i}^{3}} \right)=\left( -2i \right)\left( 1-i \right) \\\ \end{aligned}
Now multiplying the remaining two terms by using the distribution law of multiplication over the subtraction, then we will get
⇒(1−i3)=−2i+2i2\Rightarrow \left( 1-{{i}^{3}} \right)=-2i+2{{i}^{2}}
Again, substituting i2=−1{{i}^{2}}=-1 in the above equation, then we will get
⇒(1−i3)=2−2i\Rightarrow \left( 1-{{i}^{3}} \right)=2-2i

Now we will get a+bia+bi form that is 2−2i2-2i.

Note:
We can also solve this problem by using the de moivre's method.
We can write the given equation according to De moivre's theorem
(1−i)3=(2(cos⁡(−π4)+isin⁡(−π4)))3{{\left( 1-i \right)}^{3}}={{\left( \sqrt{2}\left( \cos \left( -\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{4} \right) \right) \right)}^{3}}
Now we will simplify the above expression, then
⇒(1−i)3=(2)3(cos⁡(−3π4)+isin⁡(−π4)) ⇒(1−i)3=22(−22−i22) \begin{aligned} & \Rightarrow {{\left( 1-i \right)}^{3}}={{\left( \sqrt{2} \right)}^{3}}\left( \cos \left( -\dfrac{3\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{4} \right) \right) \\\ & \Rightarrow {{\left( 1-i \right)}^{3}}=2\sqrt{2}\left( -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2} \right) \\\ \end{aligned}
Now we will simplify the above expression, then
⇒(1−i)3=−2−2i\Rightarrow {{\left( 1-i \right)}^{3}}=-2-2i
Now we will get a+bia+bi form that is 2−2i2-2i