How do I solve this equation in complex numbers 4\bar{z}^{2}-4z|z|+3=0?

Answered question

2022-01-17

How do I solve this equation in complex numbers
4z24z|z|+3=0?

Answer & Explanation

nick1337

nick1337

Expert2022-01-17Added 777 answers

Step 1
If we write:
z=|e|eiθ
and substitute, we find that we have to solve:
|z|22ieθ2e3iθ233iθ22i=34
Thus:
|z|2(sin(θ2)+icos(θ2))sin(3θ2)=38
That is, we need cos(θ2)=0. We find that necessarily we have:
θ=π
Which implies that:
sin(θ2)=1,sin(3θ2)=1
If we substitute into the last equation we find that the left hand side is negative, which is not allowed.
The given equation has no solution with zC

star233

star233

Skilled2022-01-17Added 403 answers

Step 1
1) 4z¯24z|z|+3=0
Taking the anjugate of (1) yields
2) 4z24z¯|z|+3=0
(2)(1)z2z¯2+(zz¯)|z|=0(zz¯)(z+z¯)+(zz¯)|z|=0(zz¯)(z+z¯+|z|)=0
Case 1 z=z¯
z=z¯z=aaR14a24a2+3=03=0
Thus no solutions
Step 2
Case 2 z+z¯+|z|=0
z=a+ibz+z¯=2a|z|=a2+b22a+a2+b2=0a2+b2=2aa0a2+b2=4a2b2=3a2b=±3az=a±3aia(a±3i)13=0
Thus no solutions
Sinally, the equation 4z¯24z|z|+3=0 has no solutions

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