Simplifying An Interesting Sum | Problem 305

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[@SidneiMV]

1 + i = (√2)e^(iπ/4) 1 - i = (√2)e^(-iπ/4) (1 + i)ⁿ = (√2ⁿ)e^(inπ/4) (1 - i)ⁿ = (√2ⁿ)e^(-inπ/4) (1 + i)ⁿ + (1 - i)ⁿ = (2√2ⁿ)cos(nπ/4)

[@bsmith6276]

I did it the same way. Express in polar/exponential form and then realize the sum is just twice the real part of one of the powers. I don't know why Sybermath is so reluctant to use polar/exponential form.

[@JayTemple]

What's extra funny is that KZbin offered to translate your comment to English.

[@Jim-be8sj]

Interesting one. I used the exponential and trigonometric forms of complex numbers to simplify to the alternate for you have here. It seems like a general theorem could be had for sums of complex conjugates raised to a common power.

[@samuelemorreale7510]

z=r e^it ==> z* = r e^-it z^n + z*^in = r^n (e^int + e^-int) = r^n (2cos(nt)) = 2 r^n cos(nt)

[@KrasBadan]

2•(√2)ⁿ•cos(πn/4)

[@Gezraf]

5:13 when u said there's a joke about binary: its "there are 10 types of people that know binary: those who understand it, and those who don't" binary is base2, so 10 is just 2, meaning there are two types of people: those who understand it and those who don't, granting the original logic of the sentence.

[@Qermaq]

It's kinda a "reading" joke.

[@Gezraf]

@@Qermaq yea man but it's just more to sorta indicate what he meant

[@supergamer2026]

(1+i)^n + (1-i)^n = ? we know (1+i)/(1-i) = i that means: (1+i) = i(1-i) (1+i)^n + (1-i)^n = (i(1-i))^n + (1-i)^n = i^n × (1-i)^n + (1-i)^n = (1-i)^n × (1+i^n) this works for any n

[@KipIngram]

Sure, this simplifies rather nicely. Polar form helps here: 1+i = sqrt(2)*e^(i*pi/4) 1-i = sqrt(2)*e^(-i*pi/4) (1+i)^n = 2^(n/2)*e^(i*n*pi/4) (1-i)^n = 2^(n/2)*e^(-i*n*pi/4) (1+i)^n + (1-i)^n = 2^(n/2)*[e^(i*n*pi/4) + e^(-i*n*pi/4)] (1+i)^n + (1-i)^n = 2^(1+n/2)*[e^i*n*pi/4) + e^(-i*n*pi/4)]/2 (1+i)^n + (1-i)^n = 2^(1+n/2)*cos(n*pi/4) (1+i)^n + (1-i)^n = 2*[sqrt(2)]^n*cos(n*pi/4)

[@aplusbi]

Very nice!

[@prashanthacharya5753]

Using e^i theta notation for the two terms , answer is 2 ^ (n/2 + 1) cos ( n pi / 4)

[@scottleung9587]

Cool!

[@moeberry8226]

It’s not that if n is odd then you obtain -1 and if n is even you obtain 1. n is always even since it’s a multiple of 4. The way to state it is, if n is a multiple of 4 but not a multiple of 8 then it’s -1 and if n is a multiple of 8 then it’s +1.

[@Qermaq]

Or say "congruent to 4 mod 8".

[@phill3986]

✌️👍✌️👍✌️

[@AdvaitBhalerao]

Hey I have a suggestion. How about |z|=lnz/sinz...

[@giuseppemalaguti435]

2((n,0)-((n,2)+(n,4)-(n,6)+(n,8)...

[@ILYA1991RUS_Socratus]

0?

[@aplusbi]

Sometimes

[@barthennin6088]

What an obtuse and non-intuitive solution...I like the solutions using Euler's formula much better!

[@aplusbi]

Thanks for the feedback. @KipIngram got it!

[@supergamer2026]

(1+i)^n + (1-i)^n = ? we know (1+i)/(1-i) = i that means: (1+i) = i(1-i) (1+i)^n + (1-i)^n = (i(1-i))^n + (1-i)^n = i^n × (1-i)^n + (1-i)^n = (1-i)^n × (1+i^n) this works for any n