Addition & subtraction of high power complex numbers

[LongLifeLearner]

Hello, while i tried some exercise regarding complex numbers, this question came actoss:

Use the properties z-z,*=2i Im(z) [ where z*=conjugate z] to determine the value of (sqrt(3)+i)^5-(sqrt(3)-i)^5.

I get 32i by solve it algebrally (expand the power of 5). But i have no ideal deal with the properties given. I learn De Moivre's Theorem but really have no idea how to link to this.

Hope someone can help me.Thank you.

 

[pka]

Use the properties z-z,*=2i Im(z) [ where z*=conjugate z] to determine the value of (sqrt(3)+i)^5-(sqrt(3)-i)^5.
I get 32i by solve it algebrally (expand the power of 5). But i have no ideal deal with the properties given. I learn De Moivre's Theorem but really have no idea how to link to this.

Step one change to exponential form.
\(\Theta= \arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \)
In this case \(z=\sqrt{3}+i\) so \(\Theta=\arctan\left[\dfrac{1}{\sqrt{3}}\right]~\&~|z|=2\)
Thus \(z=2\exp(\Theta i)\) Thus we get \(z^5=2^5\exp(5\Theta i)\)

 

[Dr.Peterson]

Hello, while i tried some exercise regarding complex numbers, this question came actoss:

Use the properties z-z,*=2i Im(z) [ where z*=conjugate z] to determine the value of (sqrt(3)+i)^5-(sqrt(3)-i)^5.

I get 32i by solve it algebraically (expand the power of 5). But i have no ideal deal with the properties given. I learn De Moivre's Theorem but really have no idea how to link to this.

Hope someone can help me.Thank you.

You're asking specifically, how to use the indicated property, z - z* = 2i Im(z), to evaluate that expression.

Taking z = sqrt(3) + i, the expression is (z)^5 - (z*)^5 = (z^5) - (z^5)* = 2i Im(z^5).

Now just find the imaginary part of z^5, by whatever method you like.

 

[LongLifeLearner]

You're asking specifically, how to use the indicated property, z - z* = 2i Im(z), to evaluate that expression.

Taking z = sqrt(3) + i, the expression is (z)^5 - (z*)^5 = (z^5) - (z^5)* = 2i Im(z^5).

Now just find the imaginary part of z^5, by whatever method you like.

Thank you very much. Now i saw it clearly how to apply the property given. I did solved it again with ur guidance.

 

[LongLifeLearner]

Step one change to exponential form.
\(\Theta= \arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \)
In this case \(z=\sqrt{3}+i\) so \(\Theta=\arctan\left[\dfrac{1}{\sqrt{3}}\right]~\&~|z|=2\)
Thus \(z=2\exp(\Theta i)\) Thus we get \(z^5=2^5\exp(5\Theta i)\)

This is another approach can be use to solve this question. Thanks. Just the question specifically asked to solve it by using the property given. But great to know various method

 

[pka]

Use the properties z-z,*=2i Im(z) [ where z*=conjugate z] to determine the value of (sqrt(3)+i)^5-(sqrt(3)-i)^5.

If the complex number in rectangular form, \(z=a+bi\) then \(a~\&~b\) are real numbers.
The real part of \(z\) is \(2a={z+\overline{\,z\,}}=2\Re(z)\)
The imaginary part of \(z\) is \(2bi={z-\overline{\,z\,}}=2i\Im(z)\)