How can I solve for the roots of (z+i)^2=3-4i in my complex number homework?

[HMPARTICLE]

1. The problem.

Given that z= 3-4i
Show that z^2 = 3-4i

Hence or otherwise find the roots of the equation (z+i)^2=3-4i

2. My attempt.

The first part of the problem is strait forward z^2= (2-i)(2-i) then expand to get the desired result.
Now the second part

(z+i)^2=3-4i. Becomes

z^2+ 2zi+i^2 = 3-4i

From here on I replace z with 2-i and get nowhere!

 

[maajdl]

Given that z= 3-4i
leads to z² = -7-24i
There must be a mistake in your statement.

Besides that, the equation z²=3-4i has two solutions:

z1=(2-i) and z2=-(2-i)

Solving (z+i)²=3-4i is then straightforward.

 

[HMPARTICLE]

Yes there was a mistake! It should have read... Given that z = 2-i

 

[pbuk]

So you have (2 - i)² = 3 - 4i and you are looking for roots for (x + i)² = 3 - 4i (I've changed the unknown to x to avoid confusion).

 

[HMPARTICLE]

Yes! The variable z does confuse things a bit.

 

[SammyS]

Yes! The variable z does confuse things a bit.

Don't forget, if you use x this way it is possibly a complex number !