Complex numbers: Prove sum{k=1,n}(1/4+sqrt{3}i/4)^{k-1} = (1-(1/2^n)e^{(i*pi*n)/3})/(1-(1/2)e^{(i*pi)/3})

H

HATLEY1997

Junior Member
Joined
Oct 24, 2023
Messages
59
Got on okay with part a. I understand I need to use the formula a1(1-r^2)/1-r for part b. However I am struggling with proving this as well as c. Any help or links for guidance?
 
I have went through every book and YouTube tutorial I can find and I am still struggling, can you give me a starting point?

Am I splitting the 1/4 and sqrt(3)i/4 into two seperate sigma notations as the first step?
 
For b) you need the formula for the finite geometric series which is
∑k=0n−1qk=∑k=1nqk−1=1−qn1−q \sum_{k=0}^{n-1} q^k = \sum_{k=1}^n q^{k-1}=\dfrac{1-q^{n}}{1-q}which comes from (1−q)⋅(1+q+q2+…+qn−1)=1−qn. (1-q)\cdot(1+q+q^2+\ldots+q^{n-1})=1-q^n.

Do you know how to convert the result into the notation with the e-function? (If not, then let's first see what you got and solve this issue afterward.

For c) you need the formula for the infinite geometric series which is
∑k=0∞qk=11−q \sum_{k=0}^\infty q^k=\dfrac{1}{1-q} and in your case
∑k=1∞qk=11−q−1=q1−q \sum_{k=1}^\infty q^k=\dfrac{1}{1-q} - 1=\dfrac{q}{1-q}
q=14+34i q=\dfrac{1}{4}+\dfrac{\sqrt{3}}{4} i in both cases. Remember that you divide a complex number by
1a+ib=a−ib(a−ib)(a+ib)=a−iba2+b2. \dfrac{1}{a+ i b}=\dfrac{a- i b}{(a- ib)(a+ib)}=\dfrac{a-ib}{a^2+b^2}.
 
fresh_42 said:
You forgot to tell us what you want to solve!
Click to expand...
Am I correct in saying (for b) that r=1/2e^(ipi/3) and a=1 (as substituting 1 for k into (1/4+isqrt(3)/4)^n gives me 1). Putting this into that first formula would give prove the answer required. Does r equal the answer to part a as this is what I seem to have but not sure if that is right or not
 
Let's see. I haven't calculated anything, yet.
∣q∣=∣14+34i∣=116+316=12cos⁡(φ)=14/(1/2)=1/2⟹φ=60°=π3sin⁡(φ)=34/(1/2)=32⟹φ=60°=π3q=2−1eiπ3\begin{array}{lll} |q|&=\left|\dfrac{1}{4}+\dfrac{\sqrt{3}}{4} i \right|= \sqrt{\dfrac{1}{16}+\dfrac{3}{16}}=\dfrac{1}{2}\\ \cos(\varphi) &= \dfrac{1}{4} / (1/2)= 1/2 \Longrightarrow \varphi = 60° = \dfrac{\pi}{3}\\ \sin(\varphi )&= \dfrac{\sqrt{3}}{4} / (1/2) =\dfrac{\sqrt{3}}{2} \Longrightarrow \varphi = 60° = \dfrac{\pi}{3} \\ q= 2^{-1} e^{i \frac{\pi}{3}} \end{array}
So the representation part is ok. I have trouble understanding the part with the sum. q0=1 q^0=1 and q1=q. q^1=q . I have written the formulas the way you need them, starting with k=1. k=1.

qn=(2−1eiπ3)n=2−neinπ3 q^n= \left(2^{-1} e^{i \frac{\pi}{3}}\right)^n=2^{-n}e^{i \frac{n \pi}{3}}
 
fresh_42 said:
Let's see. I haven't calculated anything, yet.
∣q∣=∣14+34i∣=116+316=12cos⁡(φ)=14/(1/2)=1/2⟹φ=60°=π3sin⁡(φ)=34/(1/2)=32⟹φ=60°=π3q=2−1eiπ3\begin{array}{lll} |q|&=\left|\dfrac{1}{4}+\dfrac{\sqrt{3}}{4} i \right|= \sqrt{\dfrac{1}{16}+\dfrac{3}{16}}=\dfrac{1}{2}\\ \cos(\varphi) &= \dfrac{1}{4} / (1/2)= 1/2 \Longrightarrow \varphi = 60° = \dfrac{\pi}{3}\\ \sin(\varphi )&= \dfrac{\sqrt{3}}{4} / (1/2) =\dfrac{\sqrt{3}}{2} \Longrightarrow \varphi = 60° = \dfrac{\pi}{3} \\ q= 2^{-1} e^{i \frac{\pi}{3}} \end{array}
So the representation part is ok. I have trouble understanding the part with the sum. q0=1 q^0=1 and q1=q. q^1=q . I have written the formulas the way you need them, starting with k=1. k=1.

qn=(2−1eiπ3)n=2−neinπ3 q^n= \left(2^{-1} e^{i \frac{\pi}{3}}\right)^n=2^{-n}e^{i \frac{n \pi}{3}}
Click to expand...

fresh_42 said:
Let's see. I haven't calculated anything, yet.
∣q∣=∣14+34i∣=116+316=12cos⁡(φ)=14/(1/2)=1/2⟹φ=60°=π3sin⁡(φ)=34/(1/2)=32⟹φ=60°=π3q=2−1eiπ3\begin{array}{lll} |q|&=\left|\dfrac{1}{4}+\dfrac{\sqrt{3}}{4} i \right|= \sqrt{\dfrac{1}{16}+\dfrac{3}{16}}=\dfrac{1}{2}\\ \cos(\varphi) &= \dfrac{1}{4} / (1/2)= 1/2 \Longrightarrow \varphi = 60° = \dfrac{\pi}{3}\\ \sin(\varphi )&= \dfrac{\sqrt{3}}{4} / (1/2) =\dfrac{\sqrt{3}}{2} \Longrightarrow \varphi = 60° = \dfrac{\pi}{3} \\ q= 2^{-1} e^{i \frac{\pi}{3}} \end{array}
So the representation part is ok. I have trouble understanding the part with the sum. q0=1 q^0=1 and q1=q. q^1=q . I have written the formulas the way you need them, starting with k=1. k=1.

qn=(2−1eiπ3)n=2−neinπ3 q^n= \left(2^{-1} e^{i \frac{\pi}{3}}\right)^n=2^{-n}e^{i \frac{n \pi}{3}}
Click to expand...
Perfect will have a look at this - thank you for your help!
 
HATLEY1997 said:
Where am I missing the 1 at the start?
Click to expand...
My mistake, I am sorry. They actually started with k=0 k=0 because the exponent in the sum was k−1 k-1 .

I gave you the wrong formula where I subtracted q0=1 q^0=1 erroneously thinking the sum was over qk. q^k. Sorry, my bad. (I blame that I watched Snooker while writing that ...)

∑k=1∞qk−1=∑k=0∞qk=11−q=1+∑k=1∞qk=1+q1−q=11−q\begin{array}{lll} \sum_{k=1}^\infty q^{k-1}=\sum_{k=0}^\infty q^{k}=\dfrac{1}{1-q}=1+\sum_{k=1}^\infty q^k =1 +\dfrac{q}{1-q}=\dfrac{1}{1-q} \end{array}
 
Last edited:
You must log in or register to reply here.
Top