600 followers problem

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

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

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

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

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

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

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

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

$X=2$

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

i^83 = -i

( 1 + i )^2 = 2 i

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

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

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

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

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,

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

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

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

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