For every real number \(\displaystyle{a}{>}{0}\). Find all

Blaze Shepherd

Blaze Shepherd

Answered question

2022-04-06

For every real number a>0. Find all complex number z, satisfying the equation z|z|+az+i=0

Answer & Explanation

anita1415snck

anita1415snck

Beginner2022-04-07Added 19 answers

Step 1
Given:
1) z|z|+az+i=0
Let z=x+iy
So, |z|=x2+y2
Substitute the above values in (1) we get
z|z|+az+0
(x+iy)(x2+y2)+a(x+iy)+i=0
xx2+y2+iyx2+y2+ax+iay+i=0
(xx2+y2+ax)+i(yx2+y2+ay+1)=0
Substituting Real part equal to zero we get
xx2+y2+ax=0
Now since z is some complex number. So, |z|0
So, x2+y20
Therefore, to satisfy the equation xx2+y2+ax=0
x=0
Substituting Imaginary part equal to zero we get
yx2+y2+ay+1=0
Substitute x=0 and solve for y we get
y(0)2+y2+ay+1=0
yy2+ay+1=0
y(y)+ay+1=0
y2+ay+1=0
Now solving the quadratic equation we get
y=a±a24(1)(1)2(1)
y=a±a242
So, the complex number z is
z=x+iy
z=0+i(a±a242)
z=i(a±a242)
Step 2
The complex numbers z satisfying the equation z|z|=az+i=0 is
z=x+iy
z=i(a±a242)
Such that
x=0 and y=a±a242

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