complex numbers

Algebra Level 2

if i z 4 + 1 = 0 iz^4 +1=0 then z can take the value

c o s π / 8 + i s i n π / 8 cosπ /8+isinπ /8 i i 1 + i / 2 1+i/\sqrt{2} 1 / 4 i 1/4i

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1 solution

Note that i = 1 i=\sqrt{-1} is an imagine number. Therefore,

i z 4 + 1 = 0 iz^4+1=0

i z 4 = 1 iz^4=-1

i z 4 = i 2 iz^4=i^2

z 4 = i z^4=i

z = cos π 8 + i sin π 8 \Rightarrow \boxed{z = \cos{\frac{\pi}{8}}+i\sin{\frac{\pi}{8}}}

To show that the above solution is correct, square z z twice as follows:

z 2 = cos 2 π 8 + 2 i sin π 8 cos π 8 sin 2 π 8 z^2 = \cos^2{\frac{\pi}{8}} +2i\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}-\sin^2{\frac{\pi}{8}}

= cos π 4 + i sin π 4 \quad = \cos{\frac{\pi}{4}} +i\sin{\frac{\pi}{4}}

Squaring z z we double the angle ( π 8 × 2 = π 4 \frac{\pi}{8}\times 2 = \frac{\pi}{4} ), similarly when we square z 2 z^2 , we get:

z 4 = cos π 2 + i sin π 2 = 0 + i = i z^4 = \cos{\frac{\pi}{2}} +i\sin{\frac{\pi}{2}}=0+i=\boxed{i}

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