Question Number 131344 by Algoritm last updated on 03/Feb/21
Answered by bluberry508 last updated on 06/Feb/21
![(1+i)=(√2)e^(((iπ)/4)+2aiπ) (a∈Z) (1+i)^n =−(1/8)(1+i) (1+i)[(1+i)^(n−1) +(1/8)]=0 ∴ (1+i)^(n−1) =−(1/8) let n−1 be m ((√2)e^(((iπ)/4)+2aiπ) )^m =(−1)((1/2))^3 =e^(iπ+2kiπ) ((1/2))^3 (k∈Z) 2^((m/2)+3) e^(iπ((m/4)+2am−1−2k)) =1 ∴(m/2)+3=0 , (m/4)+2am−1−2k = 2p (p∈Z) m=n−1=−6 thus, n=−5 then we have the following condition of p and k. 2p+2k+12a=−(5/2) but this condition must be contradictory with respect to p,k,a∈Z therefore we can′t find the value of n](Q131583.png)
$$ \\ $$$$ \\ $$$$\left(\mathrm{1}+{i}\right)=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}+\mathrm{2}{ai}\pi} \\ $$$$\left({a}\in\mathbb{Z}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}+{i}\right)^{{n}} =−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+{i}\right) \\ $$$$\left(\mathrm{1}+{i}\right)\left[\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{8}}\right]=\mathrm{0} \\ $$$$ \\ $$$$\therefore\:\:\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$${let}\:\:{n}−\mathrm{1}\:\:{be}\:\:{m} \\ $$$$ \\ $$$$\left(\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}+\mathrm{2}{ai}\pi} \right)^{{m}} =\left(−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} ={e}^{{i}\pi+\mathrm{2}{ki}\pi} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$\left({k}\in\mathbb{Z}\right) \\ $$$$ \\ $$$$\mathrm{2}^{\frac{{m}}{\mathrm{2}}+\mathrm{3}} {e}^{{i}\pi\left(\frac{{m}}{\mathrm{4}}+\mathrm{2}{am}−\mathrm{1}−\mathrm{2}{k}\right)} =\mathrm{1} \\ $$$$ \\ $$$$\therefore\frac{{m}}{\mathrm{2}}+\mathrm{3}=\mathrm{0}\:\:\:,\:\:\frac{{m}}{\mathrm{4}}+\mathrm{2}{am}−\mathrm{1}−\mathrm{2}{k}\:=\:\mathrm{2}{p}\: \\ $$$$\left({p}\in\mathbb{Z}\right) \\ $$$$ \\ $$$${m}={n}−\mathrm{1}=−\mathrm{6}\:\:\:\:{thus},\:\:{n}=−\mathrm{5} \\ $$$$ \\ $$$${then}\:\:{we}\:{have}\:{the}\:{following}\:{condition}\: \\ $$$${of}\:{p}\:{and}\:{k}. \\ $$$$\mathrm{2}{p}+\mathrm{2}{k}+\mathrm{12}{a}=−\frac{\mathrm{5}}{\mathrm{2}}\:\: \\ $$$$ \\ $$$${but}\:{this}\:{condition}\:{must}\:{be}\:{contradictory}\: \\ $$$${with}\:{respect}\:{to}\:\:\:{p},{k},{a}\in\mathbb{Z} \\ $$$$ \\ $$$${therefore}\:{we}\:{can}'{t}\:{find}\:{the}\:{value}\:{of}\:{n} \\ $$