Put z=x+iy
x+iy+α∣x+iy−1∣+2i=0
⇒x+α(x−1)2+y2+i(y+2)=0+0i
⇒y+2=0 and x+α(x−1)2+y2=0
⇒y=−2 and α2=x2−2x+5x2
Now x2−2x+5x2∈[0,45]
∴α2∈[0,45]⇒α∈[−25,25]
∴p=−25;q=25
⇒4(p2+q2)=4(45+45)=10
JEE Main 2021 — Mathematics Algebra
If the least and the largest real values of α, for which the equation z+α∣z−1∣+2i=0 (z∈Candi=−1) has a solution, are p and q respectively; then 4(p2+q2) is equal to_______.
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