What is the number of distinct solutions of the equation `z^2+|z|=0` (where z is a complex number)? (a) One (b) Two (c) Three (d) Five

Options

A. One
B. Two
C. Three
D. Five

Answer

C

Explanation

Let `z=x+iy`
`z^(2)+|z|=0`
`impliesx^(2)-y^(2)+2ixy+sqrt(x^(2)+y^(2))=0`
`impliesx^(2)y^(2)+sqrt(x^(2)+y^(2))+i2xy=0+i0`
`impliesx^(2)-y^(2)+sqrt(x^(2)+y^(2))=0" "...(1)`
`and2 xy=0impliesxy=0impliesx=0ory=0`
Now: For `y=0` in eq. (1) we get:
`x^(2)+sqrt(x^(2))=0`
`impliesx^(2)+|x|=0`
Clearly `x^(2)+|x|` will always be greater than 0 for all ` xgt0.`
Let, `x le0`
`x^(2)+|x|=0`
`impliesx^(2)-x=0impliesx(x-1)=0`
`impliesx=0or(x-1)=0`
`impliesx=0 " "(becausex le0)`
`thereforez=0`
For `x=0` in eq. (1) we get,
`-y^(2)+sqrt(y^(2))=0`
`-y^(2)+|y|=0`
`If y gt0,` then
`-y^(2)+y=0`
`implies-y^(2)|y|=0`
`impliesy=0,y=1`
`y=1" "(becausey gt0)`
`becausez=i`
If `y lt0,` then
`-y^(2)+|y|=0`
`implies-y^(2)-y=0`
`impliesy=0,y=-1`
`impliesy=1" "(becauseylt0)`
`thereforez=-i`
`therefore` There are only 3 distinct solutions.