Find the value of z complex: z^2+3z(conjugate)=4/z

[Roger.Robert]

\(\displaystyle \displaystyle Find \, the \,value \,of \,z \in \mathbb{C}^{*} \,such \,that \,\left | z \right | \neq 1 , \, Im(z) < 0 \, and \, z^{2} + 3 \bar z = \frac{4}{z}\)

 

[Deleted member 4993]

\(\displaystyle \displaystyle Find \, the \,value \,of \,z \in \mathbb{C}^{*} \,such \,that \,\left | z \right | \neq 1 , \, Im(z) < 0 \, and \, z^{2} + 3 \bar z = \frac{4}{z}\)

What do you exactly mean by:

Find the value of z complex:


Is it :

calculate the "magnitude" of \(\displaystyle \ z\)?

 

[Roger.Robert]

What do you exactly mean by:

Find the value of z complex:


Is it :

calculate the "magnitude" of \(\displaystyle \ z\)?

No,only the value.As in find the imaginary and real part of z.

 

[Deleted member 4993]

\(\displaystyle \displaystyle Find \, the \,value \,of \,z \in \mathbb{C}^{*} \,such \,that \,\left | z \right | \neq 1 , \, Im(z) < 0 \, and \, z^{2} + 3 \bar z = \frac{4}{z}\)

So start:

( x + iy)^2 + 3 * (x - iy) = 4/(x + iy)

simplify and equate the real and the imaginary part - independently to zero.

You will get two equations and two unknowns ......

 

[Roger.Robert]

So start:

( x + iy)^2 + 3 * (x - iy) = 4/(x + iy)

simplify and equate the real and the imaginary part - independently to zero.

You will get two equations and two unknowns ......

\(\displaystyle \displaystyle z^{2} + 3 \bar z = \frac{4}{z} \Longrightarrow^{*z} \, \, z^{3}+3|z|^{2}=4 \Longrightarrow \, z^{3}=4-3|z|^{2}\Longrightarrow \, z^{3}=4-3(x^{2}+y^{2})
\)

\(\displaystyle \displaystyle z^{3}=(x+iy)^{3}=x^{3}-3xy^{2}+ib(3x^{2}-y^{2}) \Longrightarrow \, x^{3}-3xy^{2}+ib(3x^{2}-y^{2})=4-3(x^{2}+y^{2}) \)

\(\displaystyle \displaystyle b(3x^{2}-y^{2})=0 \Longrightarrow \, 3x^{2}-y^{2}=0 \Longrightarrow \, y^{2}=3x^{2}\Longrightarrow^{y<0} y=\pm x \sqrt{3} \)

\(\displaystyle \displaystyle x^{3}-3xy^{2}=4-3(x^{2}+y^{2}) \Longrightarrow x^{3}-3(y^{2}(x-1)-x^{2})-4=0 \)

 

[HallsofIvy]

\(\displaystyle \displaystyle z^{2} + 3 \bar z = \frac{4}{z} \Longrightarrow^{*z} \, \, z^{3}+3|z|^{2}=4 \Longrightarrow \, z^{3}=4-3|z|^{2}\Longrightarrow \, z^{3}=4-3(x^{2}+y^{2})
\)

\(\displaystyle \displaystyle z^{3}=(x+iy)^{3}=x^{3}-3xy^{2}+ib(3x^{2}-y^{2}) \Longrightarrow \, x^{3}-3xy^{2}+ib(3x^{2}-y^{2})=4-3(x^{2}+y^{2}) \)

Although it does not affect the solution, where did that "b" come from?

\(\displaystyle \displaystyle b(3x^{2}-y^{2})=0 \Longrightarrow \, 3x^{2}-y^{2}=0 \Longrightarrow \, y^{2}=3x^{2}\Longrightarrow^{y<0} y=\pm x \sqrt{3} \)

\(\displaystyle \displaystyle x^{3}-3xy^{2}=4-3(x^{2}+y^{2}) \Longrightarrow x^{3}-3(y^{2}(x-1)-x^{2})-4=0 \)

 

[Steven G]

\(\displaystyle \displaystyle z^{2} + 3 \bar z = \frac{4}{z} \Longrightarrow^{*z} \, \, z^{3}+3|z|^{2}=4 \Longrightarrow \, z^{3}=4-3|z|^{2}\Longrightarrow \, z^{3}=4-3(x^{2}+y^{2})
\)

\(\displaystyle \displaystyle z^{3}=(x+iy)^{3}=x^{3}-3xy^{2}+ib(3x^{2}-y^{2}) \Longrightarrow \, x^{3}-3xy^{2}+ib(3x^{2}-y^{2})=4-3(x^{2}+y^{2}) \)

\(\displaystyle \displaystyle b(3x^{2}-y^{2})=0 \Longrightarrow \, 3x^{2}-y^{2}=0 \Longrightarrow \, y^{2}=3x^{2}\Longrightarrow^{y<0} y=\pm x \sqrt{3} \)

\(\displaystyle \displaystyle x^{3}-3xy^{2}=4-3(x^{2}+y^{2}) \Longrightarrow x^{3}-3(y^{2}(x-1)-x^{2})-4=0 \)

b=y, but why??

 

[Deleted member 4993]

use y^2 = 3x^2

x^3 - 3xy^2 = 4 - 3(x^2 + y^2)

x^3 - 9x^3 = 4 - 3(x^2 + 3x^2)

2x^3 - 3x^2 + 1 = 0

(x-1)(2x^2 - x - 1) = 0

And solve for 'x' hence 'y'.