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Question: How do I use DeMoivre’s theorem to find \[{\left( {1 - i} \right)^{10}}\]?...

How do I use DeMoivre’s theorem to find (1−i)10{\left( {1 - i} \right)^{10}}?

Explanation

Solution

Here, we will write the complex number in polar form by finding the value of radius rr and angle θ\theta . Then we will use DeMoivre’s theorem to find the value of the polar form. Finally, we will solve the equation and get the desired answer.
Formula used: To change complex number to polar form: z=a+biz = a + bi we will use the formula z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right), where, r=a2+b2r = \sqrt {{a^2} + {b^2}} , θ=tan⁡−1(ba)\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right) for a>0a > 0 and θ=tan⁡−1(ba)+180∘\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right) + {180^ \circ } fora<0a < 0.

Complete step by step solution:
The value given to us is
(1−i)10{\left( {1 - i} \right)^{10}}…..(1)\left( 1 \right)
The polar form of a complex number z=a+biz = a + bi is z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right).
Now comparing equation (1)\left( 1 \right) to z=a+biz = a + bi, we get
a=1a = 1
b=−1b = - 1
Now we will find rr.
Substituting a=1a = 1 and b=−1b = - 1 in the formula r=a2+b2r = \sqrt {{a^2} + {b^2}} , we get
So,
r=(1)2+(−1)2r = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}}
Simplifying the expression, we get
⇒r=2\Rightarrow r = \sqrt 2
Now we can see that a>0a > 0 so θ=tan⁡−1(ba)\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right).
Substituting the values in the above formula, we get
θ=tan⁡−1(−11)\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{1}} \right)
⇒θ=tan⁡−1(−1)\Rightarrow \theta = {\tan ^{ - 1}}\left( { - 1} \right)
Using the trigonometric table, we get
⇒θ=−π4\Rightarrow \theta = - \dfrac{\pi }{4}
Now substituting all the obtained values inz=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right) we get,
(1−i)10=(2(cos⁡(−π4)+isin⁡(−π4)))10{\left( {1 - i} \right)^{10}} = {\left( {\sqrt 2 \left( {\cos \left( { - \dfrac{\pi }{4}} \right) + i\sin \left( { - \dfrac{\pi }{4}} \right)} \right)} \right)^{10}}
Using DeMoivre’s Theorem where (cos⁡(x)+isin⁡(x))n=cos⁡(nx)+isin⁡(nx){\left( {\cos \left( x \right) + i\sin \left( x \right)} \right)^n} = \cos \left( {nx} \right) + i\sin \left( {nx} \right) we get,
⇒(1−i)10=(2)10(cos⁡(−10×π4)+isin⁡(−10×π4))\Rightarrow {\left( {1 - i} \right)^{10}} = {\left( {\sqrt 2 } \right)^{10}}\left( {\cos \left( { - 10 \times \dfrac{\pi }{4}} \right) + i\sin \left( { - 10 \times \dfrac{\pi }{4}} \right)} \right)
Simplifying the expression, we get
⇒(1−i)10=32(cos⁡(−5π2)+isin⁡(−5π2))\Rightarrow {\left( {1 - i} \right)^{10}} = 32\left( {\cos \left( {\dfrac{{ - 5\pi }}{2}} \right) + i\sin \left( { - \dfrac{{5\pi }}{2}} \right)} \right)
We know that cos⁡(−5π2)=0\cos \left( {\dfrac{{ - 5\pi }}{2}} \right) = 0 and sin⁡(−5π2)=−1\sin \left( {\dfrac{{ - 5\pi }}{2}} \right) = - 1.
Using the values of angle, we get
(1−i)10=−32i{\left( {1 - i} \right)^{10}} = - 32i

So value of (1−i)10{\left( {1 - i} \right)^{10}} is −32i - 32i

Note:
DeMoivre’s theorem states that for any real number xx and integer nn the below value holds true: (cos⁡(x)+isin⁡(x))n=cos⁡(nx)+isin⁡(nx){\left( {\cos \left( x \right) + i\sin \left( x \right)} \right)^n} = \cos \left( {nx} \right) + i\sin \left( {nx} \right).
For using DeMoivre’s theorem on any complex number we should change it in its polar form then only we can use the theorem. It is used to compute the powers of complex numbers. As we know when we multiply a complex number by itself then the absolute value is squared and the argument is multiplied by 2. This concept is used to form a theorem that can be applied to any complex number with a higher power to reduce the calculation and get the result easily. The polar form is another way to represent a complex number apart from the rectangular form.