# Thread: Find the smallest integer n

1. ## Find the smallest integer n

Question: Find the smallest integer n for which $\left(\frac{1+i}{1-i} \right)^n = 1$

2. Start by converting to polar coordinates:

$\left(\frac{1+i}{1-i} \right)^n = (\frac{\sqrt 2 \arg 45}{\sqrt 2 \arg -45})^n=\frac{(\sqrt 2)^n \arg (45*n)}{(\sqrt 2)^n \arg (-45*n)}=1 \arg (90*n)$

Now we can see that the solutions are n=...-12,-8,-4,0,4,8,12 ...

3. ## I am still unable to understand

Originally Posted by Kiwi_Dave
Start by converting to polar coordinates:

$\left(\frac{1+i}{1-i} \right)^n = (\frac{\sqrt 2 \arg 45}{\sqrt 2 \arg -45})^n=\frac{(\sqrt 2)^n \arg (45*n)}{(\sqrt 2)^n \arg (-45*n)}=1 \arg (90*n)$

Now we can see that the solutions are n=...-12,-8,-4,0,4,8,12 ...

Thanks for the post but i still cannot understand

4. Zorro

Can you answer "convert 1+i to polar coordinates". If you can show the notation that you are familiar with then it will be easier to answer your question.

I would say $1+i=\sqrt 2 \arg {45}$ but you may have been taught to write it in a different way.

5. ## My work ,please let me know what i have done wrongly

Originally Posted by Kiwi_Dave
Zorro

Can you answer "convert 1+i to polar coordinates". If you can show the notation that you are familiar with then it will be easier to answer your question.

I would say $1+i=\sqrt 2 \arg {45}$ but you may have been taught to write it in a different way.

$\left( \frac{1 + i}{1 - i} \right)^n$ = $\frac{cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4})}{cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4})}$

I am stuck here .....what should i do now??????

6. $\left( \frac{1 + i}{1 - i} \right)^n$ = $\frac{cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4})}{cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4})}$ = $\frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}{(cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}$ = $\frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))^2}{cos^2(\frac{n \pi}{4}) + sin^2 (\frac{n \pi}{4})}$ = $\frac{(cos(\frac{n \pi}{2}) + i sin (\frac{n \pi}{2}))}{1}$

Now you want this to equal 1 so the complex part must be zero. That is

$sin (\frac{n \pi}{2})=0$, this occurs if and only if n is even.

Now find the EVEN values of n such that:

$cos (\frac{n \pi}{2})=1$

7. ## smallest number

Originally Posted by Kiwi_Dave
$\left( \frac{1 + i}{1 - i} \right)^n$ = $\frac{cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4})}{cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4})}$ = $\frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}{(cos(\frac{n \pi}{4}) - i sin (\frac{n \pi}{4}))(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))}$ = $\frac{(cos(\frac{n \pi}{4}) + i sin (\frac{n \pi}{4}))^2}{cos^2(\frac{n \pi}{4}) + sin^2 (\frac{n \pi}{4})}$ = $\frac{(cos(\frac{n \pi}{2}) + i sin (\frac{n \pi}{2}))}{1}$

Now you want this to equal 1 so the complex part must be zero. That is

$sin (\frac{n \pi}{2})=0$, this occurs if and only if n is even.

Now find the EVEN values of n such that:

$cos (\frac{n \pi}{2})=1$

I beleive the smallest the number would n = 0
Am i correct ???

8. Hello, zorro!

Find the smallest integer $n$ for which: . $\left(\frac{1+i}{1-i} \right)^n \:=\: 1$

We have: . $\frac{(1+i)^n}{(1-i)^n} \:=\:1$

Multiply by $(1-i)^n\!:\quad (1+i)^n \:=\:(1-i)^n$

Multiply by $(1 + i)^n\!:\quad (1+i)^n\cdot(1+i)^n \:=\:(1+i)^n(1-i)^n$

. . . . . $(1+i)^{2n} \:=\:\bigg[(1+i)(1-i)\bigg]^n \;=\;\left(1 - i + i - i^2\right)^n$

And we have: . $(1+i)^{2n} \:=\:2^n$

By inspection: . $n \,=\,0$

9. n=-4000 is an integer less than 0 and $cos(\frac{-4000 \pi}{2})=1$ so -4000 is also a solution. There is no unique smallest solution because for any solution n, n-4 is also a solution.