Evaluate

$\large{ \frac { { (1+i) }^{ 85 } }{ { (1-i) }^{ 83 } }}$

**
Hint
**
: There's an extremely fast way to solve this if you know why

$\large 1+i = \sqrt{2}e^{i{\pi}/{4}}$

1
2
-i
2i

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Hey , I am amazed by seeing your profile page , you are able to solve almost every problem on Brilliant.org. You recently solved my problem too i.e This problem Can you please provide a solution to it , I am trying it for almost 10 days , I will surely upvote and will be very thankful to learn from such a brilliant child. @abdulrahman khaled

U Z
- 6 years, 5 months ago

Used the same method but made (1-i)^83 as ((1-i)^2)^42 * (1-i)^-1

:D Nice question

Farouk Yasser
- 6 years, 5 months ago

Euler is better with higher powers! See @Chew-Seong Cheong 's solution. Btw its a nice idea to answer it "2.000" to make it decimal question!!

Pranjal Jain
- 6 years, 5 months ago

Great... now i understand

Andro Dellosa
- 6 years, 5 months ago

Very good! But is it necessary to let it equal X?

Whitney Clark
- 4 years, 7 months ago

$X = \dfrac {(1+i)^{85}}{(1-i)^{83}} = \dfrac {(\sqrt{2}e^{\frac {\pi}{4}i})^{85}}{(\sqrt{2}e^{\frac {-\pi}{4}i})^{83}} = (\sqrt{2})^{85-83} e^{\frac {\pi}{4}(85+83)i} = 2 e^{42\pi i} = \boxed{2}$

**
Reference:
**
Euler's formula
$e^{\theta i} = \cos \theta + i \sin \theta$
.

37 Helpful
3 Interesting
0 Brilliant
0 Confused

can u please tell me which formula u have used

Guru Prasaadh
- 6 years, 5 months ago

Log in to reply

a complex number can be written in the form as shown- It's know as euler's representation i.e $x + iy = r(\cos\theta + i\sin\theta) = re^{i\theta}$

U Z
- 6 years, 5 months ago

Log in to reply

oh thank you

Guru Prasaadh
- 6 years, 5 months ago

how did that 2 e power 42i become 2?

Guru Prasaadh
- 6 years, 5 months ago

Log in to reply

@Guru Prasaadh – $2 e^{42i\pi} = 2(cos42 \pi + isin42\pi) = 2$

U Z
- 6 years, 5 months ago

That's some sort of trigonometric form or De Moivre's Theorem is it? Looks the same when I saw the theorem in a trigonometry book.

Marc Vince Casimiro
- 6 years, 5 months ago

Guru, Megh is right. It is called Euler's identity . Euler is a great mathematician who linked imaginary numbers to the real world.

Chew-Seong Cheong
- 6 years, 5 months ago

I'm pretty sure he used Euler's Formula.

Vikram Sarkar
- 10 months, 3 weeks ago

$X= \frac{(1+i)^{84} \times (1+i)}{(1-i)^{83}}$

Multiplying by $\frac{(1-i)}{(1-i)}$ :

$X= \frac{(1+i)^{84} \times (1+i) \times (1-i)}{(1-i)^{84}}$

$X= \frac{((1+i)^2)^{42} \times 2}{((1-i)^2)^{42}}$

$X= \frac{(2i)^{42} \times 2}{(-2i)^{42}}$

$X=2$

9 Helpful
0 Interesting
0 Brilliant
0 Confused

Nice, I think it is the most practical way for the solution.

Abdulmennân Tor
- 5 years, 3 months ago

( 1 + i )/( 1 + i) = i

i^83 = -i

( 1 + i )^2 = 2 i

X = (2 i)(-i) = 2

3 Helpful
0 Interesting
0 Brilliant
0 Confused

it is clear that (1+i)/(1-i) is equal to i. So X is i^83 * (1+i)^2. So X is 2

2 Helpful
0 Interesting
0 Brilliant
0 Confused

1-i= -i(1+i) Substitute and simplify (1+i)^2/(-i)^83=2i/(-i)^3=2i/i=2

0 Helpful
0 Interesting
0 Brilliant
0 Confused

solution : x = {[(1 + i)^85]/[(1 - i)^83]} = 2 so X = 2 is the answer.

0 Helpful
0 Interesting
0 Brilliant
0 Confused

What I did was rewrite the complex numbers in trigonometric form. If you are to plot 1+i and 1-i on the complex plane, you get two 45, 45, 90 triangles. One of them is in the first quadrant, and the second is in the fourth. The legs of both triangles are 1, so the hypotenuse (or the absolute value of each number) is sqrt(2). Because the angle above the x-axis is 45 for the first number and -45 for the second, you can rewrite the problem as follows:

$X = \frac{(\sqrt{2}cis(45))^{85}}{(\sqrt{2}cis(-45))^{83}}$

Rewrite as

$X = \frac{\sqrt{2}^{85}}{\sqrt{2}^{83}} \times \frac{cis(45)^{85}}{cis(-45)^{83}}$

The first fraction can be simplified to $\sqrt{2}^{85-83} = \sqrt{2}^{2} = {2}$ To simplify the second, we need to use De Moivre's Theorem

$\frac{cis(45)^{85}}{cis(-45)^{83}} = \frac{cis(45 \times 85)}{cis(-45 \times 83)} = cis((45 \times 85) - (-45 \times 83)) = cis( 45 \times (85 --83)) = cis(45 \times(85+83)) = cis(45 \times 168) = cis (7560)$

Luckily, 7560 happens to be 21 revolutions. So if we're talking angles, 7560 is equivalent to 0.

So,

$X = \frac{\sqrt{2}^{85}}{\sqrt{2}^{83}} \times \frac{cis(45)^{85}}{cis(-45)^{83}} = 2cis(0)$

Remember that cis means $\cos \theta + i \sin \theta$

Therefore,

$cis(0) = cos(0) + isin(0) = 1 + i(0) = 1 + 0 = 1$

Therefore,

$2cis(0) = 2(1) = \boxed{2}$

0 Helpful
0 Interesting
0 Brilliant
0 Confused

This approach is very artistic. I solved it in many different ways (using Algebra) but using this (trigonometric transformation).

Roman Frago
- 6 years, 5 months ago

X=(((-i+1)/i)^85)/((1-i)^83) X= ((1-i)^2)/-i X= 2i/i

0 Helpful
0 Interesting
0 Brilliant
0 Confused

X=(1+i)^2 * (1+i/1-i)^83 X=(1+i)^2 * (i)^83 X=(2i) * (i)^3 X=(2i) * (-i) X= 2* -(-1) X=2

0 Helpful
0 Interesting
0 Brilliant
0 Confused

X=(1+i)^2 * (1+i/1-i)^83 X=(1+i)^2 * (i)^83 X=(2i) * (i)^3 X=(2i) * (-i) X= 2* -(-1) X=2

0 Helpful
0 Interesting
0 Brilliant
0 Confused

×

Problem Loading...

Note Loading...

Set Loading...

Let the given expression be $X$ . Now solving it further: $X=\frac { { (1+i) }^{ 84 }(1+i) }{ { (1-i) }^{ 82 }(1-i) } \\ X=\frac { { ({ (1+i) }^{ 2 }) }^{ 42 }(1+i) }{ { ({ (1-i) }^{ 2 }) }^{ 41 }(1-i) } \\ X=\frac { { (2i) }^{ 42 }(1+i) }{ { (-2i) }^{ 41 }(1-i) } \\ X=\frac { -2i(1+i) }{ (1-i) } \\ X=\frac { -2i-2{ i }^{ 2 } }{ (1-i) } \\ X=\frac { -2i+2 }{ (1-i) } \\ X=\frac { 2(1-i) }{ (1-i) } \\ X=\boxed { 2 }$