# Complex number question

1. Jul 16, 2006

### 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.

2. Jul 16, 2006

### nazzard

Hello Jack,

do you know the following representation of a complex number $z$ in the complex plane?

$$z=x+iy=r(\cos\phi+i\,\sin\phi)=r e^{i\phi}$$

Here's a sketch to clarify what $r$ and $\phi$ are meant to be.

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

Regards,

nazzard

Last edited by a moderator: May 2, 2017
3. Jul 17, 2006

### interested_learner

A couple of useful formulas:

if
$$K e^{i\psi} = a + ib \quad then \quad K = \sqrt{a^2 + b^2} \quad \psi = tan^{-1} b/a \quad$$
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

Last edited: Jul 17, 2006
4. Jul 17, 2006

### nazzard

Thank you tony. I want to point out that one has to be very careful with using arctan to calculate $$\phi$$, 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

5. Jul 17, 2006

### Office_Shredder

Staff Emeritus
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)

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