Help Me Solve $\text{Im}(z^3)$ in Rectangular Form

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SUMMARY

The discussion focuses on finding the imaginary part of the expression $\text{Im}(z^3)$ where $z = x + jy$. The key conclusion is that the imaginary part simplifies to $(3x^2y - y^3)$. This result is derived from the expansion of $z^3$ and identifying the terms containing the imaginary unit $j$. The final answer confirms that the imaginary component is indeed $3x^2y - y^3$.

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  • Familiarity with polynomial expansion and the binomial theorem.
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Drain Brain
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I don't how to proceed further

Let $z=x+jy$ find in rectangular form:

1. $\text{Im}(z^3)$

this is where I can get to,

$\text{Im}(x^3+3x^2jy+3xj^2y^2+j^3y^3)=\text{Im}(x^3+3x^2jy-3xy^2-jy^3)$

please help me.
 
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Drain Brain said:
I don't how to proceed further

Let $z=x+jy$ find in rectangular form:

1. $\text{Im}(z^3)$

this is where I can get to,

$\text{Im}(x^3+3x^2jy+3xj^2y^2+j^3y^3)=\text{Im}(x^3+3x^2jy-3xy^2-jy^3)$

please help me.

So what are the terms which have a "j" part?
 
Drain Brain said:
I don't how to proceed further

Let $z=x+jy$ find in rectangular form:

1. $\text{Im}(z^3)$

this is where I can get to,

$\text{Im}(x^3+3x^2jy+3xj^2y^2+j^3y^3)=\text{Im}(x^3+3x^2jy-3xy^2-jy^3)$

please help me.

Hi Drain Brain! :)

You're already there:
$$\text{Im}(x^3+3x^2jy-3xy^2-jy^3) = \text{Im}((x^3-3xy^2)+(3x^2y-y^3)j)
= (3x^2y-y^3)$$
 
they are $3x^2yj-j^3$ is this the answer?
 
Drain Brain said:
they are $3x^2yj-j^3$ is this the answer?

no .as I like serena mentioned it is
$3x^2y- y^3$ because imaginary part of $x+jy$ is $y $
 

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