Q.
The complex number which satisfies the equation z+2∣z+1∣+i=0 is
1807
199
Complex Numbers and Quadratic Equations
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Solution:
Since z+2∣z+1∣+i=0 ∴x+i(y+1)+2∣x+iy+1∣=0 ∴y+1=0 (∵∣x+iy+1∣ is real ) ∴y=−1 ∴x+2∣x−i+1∣=0 ⇒x2=2[(x+1)2+1] =2(x2+2x+2) ∴x2+4x+4=0 ⇒(x+2)2=0
or x=−2 ∴z=−2−i.