$$\\$$$$\\$$$$\left(\mathrm{1}+{i}\right)=\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}+\mathrm{2}{ai}\pi} \\$$$$\left({a}\in\mathbb{Z}\right) \\$$$$\\$$$$\left(\mathrm{1}+{i}\right)^{{n}} =−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+{i}\right) \\$$$$\left(\mathrm{1}+{i}\right)\left[\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{8}}\right]=\mathrm{0} \\$$$$\\$$$$\therefore\:\:\left(\mathrm{1}+{i}\right)^{{n}−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{8}} \\$$$${let}\:\:{n}−\mathrm{1}\:\:{be}\:\:{m} \\$$$$\\$$$$\left(\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}+\mathrm{2}{ai}\pi} \right)^{{m}} =\left(−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} ={e}^{{i}\pi+\mathrm{2}{ki}\pi} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} \\$$$$\left({k}\in\mathbb{Z}\right) \\$$$$\\$$$$\mathrm{2}^{\frac{{m}}{\mathrm{2}}+\mathrm{3}} {e}^{{i}\pi\left(\frac{{m}}{\mathrm{4}}+\mathrm{2}{am}−\mathrm{1}−\mathrm{2}{k}\right)} =\mathrm{1} \\$$$$\\$$$$\therefore\frac{{m}}{\mathrm{2}}+\mathrm{3}=\mathrm{0}\:\:\:,\:\:\frac{{m}}{\mathrm{4}}+\mathrm{2}{am}−\mathrm{1}−\mathrm{2}{k}\:=\:\mathrm{2}{p}\: \\$$$$\left({p}\in\mathbb{Z}\right) \\$$$$\\$$$${m}={n}−\mathrm{1}=−\mathrm{6}\:\:\:\:{thus},\:\:{n}=−\mathrm{5} \\$$$$\\$$$${then}\:\:{we}\:{have}\:{the}\:{following}\:{condition}\: \\$$$${of}\:{p}\:{and}\:{k}. \\$$$$\mathrm{2}{p}+\mathrm{2}{k}+\mathrm{12}{a}=−\frac{\mathrm{5}}{\mathrm{2}}\:\: \\$$$$\\$$$${but}\:{this}\:{condition}\:{must}\:{be}\:{contradictory}\: \\$$$${with}\:{respect}\:{to}\:\:\:{p},{k},{a}\in\mathbb{Z} \\$$$$\\$$$${therefore}\:{we}\:{can}'{t}\:{find}\:{the}\:{value}\:{of}\:{n} \\$$