Solve \(\displaystyle{z}^{{{2}}}+{\left({2}+{2}{i}\right)}{z}+{i}={0}\)

markush35q

markush35q

Answered question

2022-03-23

Solve z2+(2+2i)z+i=0

Answer & Explanation

Kamora Campbell

Kamora Campbell

Beginner2022-03-24Added 13 answers

Step 1: Solve
We have,
z=x+iy
Therefore,
z2+(2+2i)z+i=0
(x+iy)2+(2+2i)(x+iy)+i=0
x2+2xyiy2+2x+2yi=2ξ2y+i=0
x2y2+2x2y+i(2xy+2x+2y)=0
x2+2x+11y22y+11+i(2xy+2x+2y)=0
(x+1)21(y+1)2++i(2xy+2x+2y)=0
(x+1)2(y+1)2+i(2xy+2x+2y)=0

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