Solve the equation.

Attachment 19990

I don't know how to start and what to do here.

Maybe I start with inserting the square root of minus one, or changing Re(z) = 3, and Im(z) = 1.

I really don't know.

Please for some advice and help.

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- Dec 6th 2010, 12:26 PMNforceEquation with complex numbers
Solve the equation.

Attachment 19990

I don't know how to start and what to do here.

Maybe I start with inserting the square root of minus one, or changing Re(z) = 3, and Im(z) = 1.

I really don't know.

Please for some advice and help. - Dec 6th 2010, 12:31 PMtonio
- Dec 6th 2010, 12:32 PMTheEmptySet
This should get you started any complex number $\displaystyle z$ can be written in the form

$\displaystyle z=x+iy$ then $\displaystyle \text{Re}(z)=x$ and$\displaystyle \text{Im}(z)=y$

and note that two complex numbers are equal if and only if there real and imaginary parts are equal. for example

$\displaystyle z=3+4i$ if and only if $\displaystyle x=3 \text{ and } y=4$

Try this and post again if you get stuck. - Dec 6th 2010, 12:33 PMPlato
That is just $\displaystyle (2xy)i=(x+3)+(-y+1)i.$

- Dec 6th 2010, 01:07 PMNforce
I still can't see what are we doing. All those x,y and i confused me. At the "standard" equation you just have one variable.

- Dec 6th 2010, 01:14 PMPlato
- Dec 8th 2010, 12:39 AMNforce
Now I understand, I have just a linear equation, I need to find x, and y and then insert them back to see if left side is equal to right. And I have to set the real parts of equation to equal and same for imaginary part

Am I right?

I don't have an instructor, this forum is my instructor. - Dec 8th 2010, 03:12 AMPlato
- Dec 9th 2010, 07:07 AMNforce
I apologize, but I understand now, thanks to this forum and you guys.