Question

complex plane

Answer 1

\(\huge (\frac{1 + i}{1-i})^{2718}\)

Answer 2

angle = \(45-(-45) = 90\)

so thats simply \(\large e^{i\frac{\pi}{2}*2718}\)

Answer 3

consider \(1+i=(1-i)^*\) hence $$\frac1{1-i}=\frac{(1-i)^*}{\sqrt2}=\frac{1+i}{\sqrt2}$$ so $$\frac{1+i}{1-i}=\frac1{\sqrt2}(1+i)^2=\sqrt2i$$

Answer 4

ergo \((\sqrt2i)^{2718}=-2^{1359}\)

Answer 5

or it might actually have been \(2\) not \(\sqrt2\) in which case i meant \(i^{2718}=-1\) since \(2718=2716+2\equiv2\pmod4\)

Answer 6

One could actually just perform the division too,\[\frac{ 1+i }{ 1-i } \equiv \frac{(1+i)^2}{1^2 - i^2} = i\]

Answer 7

looks irishboy cooked up the problem from \(e^{i\pi}=-1\)

Answer 8

\[\left\{ \frac{ 1+\iota }{ 1-\iota } \times \frac{ 1+\iota }{ 1+ \iota } \right\}^{2718}\]
\[=\left( \frac{ 1+\iota ^2+2 \iota }{ 1-\iota ^2 } \right)^{2718}\]
\[=(\frac{ 1-1+2 \iota }{ 1-(-1) })^{2718}\]
\[=\left( \iota ^2 \right)^{1359}=\left( -1 \right)^{1359}=-1\]

Answer 9

thanks everyone for the help!