## IrishBoy123 one year ago complex plane

1. IrishBoy123

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

2. ganeshie8

angle = $$45-(-45) = 90$$ so thats simply $$\large e^{i\frac{\pi}{2}*2718}$$

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. ganeshie8

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

8. anonymous

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

9. IrishBoy123

thanks everyone for the help!