Complex Number Question: Solving for Square Roots and Equations

[jack1234]

Hi, how to solve this question?

Find the square roots fo the complex number -40-42i.
Hence
(i) Find the square roots of the complex number 40+42i,
(ii) solve the equation (z+1)^2 + 160 + 168i = 0 for all complex roots.

I don't know how to start solving this question.

 

[nazzard]

Hello Jack,

do you know the following representation of a complex number [itex]z[/itex] in the complex plane?

[tex]z=x+iy=r(\cos\phi+i\,\sin\phi)=r e^{i\phi}[/tex]

Here's a sketch to clarify what [itex]r[/itex] and [itex]\phi[/itex] are meant to be.

http://upload.wikimedia.org/wikipedia/en/c/c2/Complex.png

Regards,

nazzard

 

[interested_learner]

A couple of useful formulas:

if
[tex]
K e^{i\psi} = a + ib \quad
then \quad
K = \sqrt{a^2 + b^2} \quad
\psi = tan^{-1} b/a \quad
[/tex]
That way you can convert from one form to another. Since taking a square root is easy in the alternative form, you should have no problem.

This is what nazzard said, but maybe in a manner that is a bit clearer to a beginner in complex numbers. It is really just the application of the pythagorean theorum to the chart above. Compare what I wrote to nazzard's picture. You'll see,

Tony

 

[nazzard]

Thank you tony. I want to point out that one has to be very careful with using arctan to calculate [tex]\phi[/tex], the so called complex argument of z. Remember: case differentiation for different values of b and a (or y and x in my post).

Regards,

nazzard

 

[Office_Shredder]

Of course, there's a (possibly) more intuitive and direct way of doing it.

If sqrt(-40-42i) = (a+bi), then just solve for a and b by squaring both sides (you should be able to fnid two independent equations to break it up into)