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Question: How do you use de moivre’s theorem to simplify \({{(1+i)}^{5}}\) ?...

How do you use de moivre’s theorem to simplify (1+i)5{{(1+i)}^{5}} ?

Explanation

Solution

We have to simplify a complex number by using de moivre’s theorem. The de moivre theorem states that if z=r(cos⁡θ+isin⁡θ)z=r(\cos \theta +i\sin \theta ) is a complex number, then zn=[r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ){{z}^{n}}={{[r(\cos \theta +i\sin \theta )]}^{n}}={{r}^{n}}(\cos n\theta +i\sin n\theta ) , where θ\theta is the angle, r is the radius and n is the exponent. So, we start solving this problem, by changing the given question in the form of if z=r(cos⁡θ+isin⁡θ)z=r(\cos \theta +i\sin \theta ), and then taking exponent 5 on both sides. Then, apply the quadrant rule and make the necessary calculations. After that take LCM in the equation and then apply the distributive property a(b−c)=ab−aca(b-c)=ab-ac to get the required solution.

Complete answer:
The complex number is given in the question: (1+i)5{{(1+i)}^{5}} ----------- (1)

As we know, the de moivre’s theorem states that,
If z=r(cos⁡θ+isin⁡θ)z=r(\cos \theta +i\sin \theta ) is a complex number, ---------- (2)
Then, zn=[r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ){{z}^{n}}={{[r(\cos \theta +i\sin \theta )]}^{n}}={{r}^{n}}(\cos n\theta +i\sin n\theta ) ---------- (3)

So firstly, we will convert equation (1) in the form of equation (2), for that we need to find the value of polar forms that is, r, cos⁡θ, and sin⁡θ\cos \theta ,\text{ and }\sin \theta .
As we know, r is the radius, therefore formula to calculate r: r=a2+b2r=\sqrt{{{a}^{2}}+{{b}^{2}}} , where a and b are the real numbers of equation (1), thus we get
r=a2+b2 ⇒r=12+12 ⇒r=1+1 \begin{aligned} & r=\sqrt{{{a}^{2}}+{{b}^{2}}} \\\ & \Rightarrow r=\sqrt{{{1}^{2}}+{{1}^{2}}} \\\ & \Rightarrow r=\sqrt{1+1} \\\ \end{aligned}
⇒r=2\Rightarrow r=\sqrt{2} ----------------- (4)

Also, for the value of angle θ\theta , we know that θ=tan⁡−1(ba)\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right) , where a and b are the real number of equation (1), we get
⇒θ=tan⁡−1(ba) ⇒θ=tan⁡−1(11) ⇒θ=tan⁡−1(1) \begin{aligned} & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right) \\\ & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{1}{1} \right) \\\ & \Rightarrow \theta ={{\tan }^{-1}}(1) \\\ \end{aligned}
As, we see from the complex number, that both the real part and imaginary part are positive, therefore it lies in the first quadrant of the cartesian plane , therefore
⇒θ=π4\Rightarrow \theta =\dfrac{\pi }{4}
Thus, the value of cos⁡θ and sin⁡θ\cos \theta \text{ and }\sin \theta are:
cos⁡θ=cos⁡(π4)  \begin{aligned} & \cos \theta =\cos \left( \dfrac{\pi }{4} \right) \\\ & \\\ \end{aligned} ------------ (5)

sinθ=sin⁡(π4)  \begin{aligned} & \text{sin}\theta =\sin \left( \dfrac{\pi }{4} \right) \\\ & \\\ \end{aligned} ------------- (6)
Now, we will substitute the equation (4), (5), and (6) in equation (2), to get a trigonometric form, we get
⇒z=2(cos⁡π4+i.sin⁡π4)\Rightarrow z=\sqrt{2}\left( \cos \dfrac{\pi }{4}+i.\sin \dfrac{\pi }{4} \right) ----------- (7)

Now, we will use de moivre’s theorem in equation (7), to get a more simplified answer, that is
⇒zn=[r(cos⁡θ+isin⁡θ)]n ⇒z5=[2(cos⁡π4+isin⁡π4)]5 \begin{aligned} & \Rightarrow {{z}^{n}}={{[r(\cos \theta +i\sin \theta )]}^{n}} \\\ & \Rightarrow {{z}^{5}}={{\left[ \sqrt{2}\left( \cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4} \right) \right]}^{5}} \\\ \end{aligned}
zn=rn(cos⁡nθ+isin⁡nθ){{z}^{n}}={{r}^{n}}(\cos n\theta +i\sin n\theta )
⇒z5=(2)5(cos⁡5π4+isin⁡5π4)\Rightarrow {{z}^{5}}={{(\sqrt{2})}^{5}}\left( \cos \dfrac{5\pi }{4}+i\sin \dfrac{5\pi }{4} \right)
Therefore, doing further calculations, we get
⇒z5=(2)5(cos⁡(π+π4)+isin⁡(π+π4))\Rightarrow {{z}^{5}}={{(\sqrt{2})}^{5}}\left( \cos \left( \pi +\dfrac{\pi }{4} \right)+i\sin \left( \pi +\dfrac{\pi }{4} \right) \right)
Now, we know that π+θ\pi +\theta lies in the third quadrant, and both sin⁡θ\sin \theta and cos⁡θ\cos \theta are negative in the third quadrant, therefore we get
⇒z5=(2)5(−cos⁡(π4)−isin⁡(π4)) ⇒z5=(2)5(−(12)−i(12)) ⇒z5=(2)5(−12−i.12) \begin{aligned} & \Rightarrow {{z}^{5}}={{(\sqrt{2})}^{5}}\left( -\cos \left( \dfrac{\pi }{4} \right)-i\sin \left( \dfrac{\pi }{4} \right) \right) \\\ & \Rightarrow {{z}^{5}}={{(\sqrt{2})}^{5}}\left( -\left( \dfrac{1}{\sqrt{2}} \right)-i\left( \dfrac{1}{\sqrt{2}} \right) \right) \\\ & \Rightarrow {{z}^{5}}={{\left( \sqrt{2} \right)}^{5}}\left( \dfrac{-1}{\sqrt{2}}-i.\dfrac{1}{\sqrt{2}} \right) \\\ \end{aligned}
Thus, taking LCM and on further solving we get

& \Rightarrow {{z}^{5}}={{\left( \sqrt{2} \right)}^{4}}(-1-i) \\\ & \Rightarrow {{z}^{5}}=4.(-1-i) \\\ & \\\ \end{aligned}$$ On, applying the distributive property $a(b-c)=ab-ac$ in the above equation, we get $\Rightarrow {{z}^{5}}=-4-4i$ $\Rightarrow {{(1+i)}^{5}}=-4-4i$ , using de moivre’s theorem. **Note:** Always complete all the steps and make the necessary calculations to avoid mistakes. Always mention the de moivre’s theorem. One of the alternative methods to solve this problem is by using the exponential formula, that is compare the given equation with this formula $(a+ib)=\sqrt{{{a}^{2}}+{{b}^{2}}}.{{e}^{i\theta }}$ where, ${{e}^{i\theta }}=\cos \theta +i\sin \theta $ $\theta ={{\tan }^{-1}}\dfrac{b}{a}$ , to get your required solution.