AP EAPCET 18 May 2024 Shift 1 Solved Paper

Given the equation z3+z=0, where z=x+iy is a complex number, we can find the distinct solutions as follows:
Rewrite the Equation:
Start with:
z3+z=0
This implies:
z3=−z
Equate Magnitudes:
Taking the magnitude of both sides, we have:
|z3|=|−z|
This simplifies to:
|z|3=|z|
Since |z|=|z|, the equation becomes:
|z|3=|z|
Solve for Magnitude:
This equation leads to:
|z|(|z|2−1)=0
Therefore, either |z|=0 or |z|2=1.
Consider Each Case:
If |z|=0, then z=0. This gives one distinct solution: z=0.
If |z|2=1, then |z|=1 and consequently zz=1. This implies:
z=‌

1

z

Substitute Back:
Substituting z=‌

1

z

into the original equation:
z3+‌

1

z

=0
Multiply through by z to clear the fraction:
z4+1=0
Solve z4+1=0 :
This equation has four roots:
z=ei(π∕4),ei(3π∕4),ei(5π∕4),ei(7π∕4)
Count Distinct Solutions:
Including z=0, we have:

One solution for z=0
Four solutions from z4+1=0
Therefore, the number of distinct solutions to the equation is 5 .