complex equation

[newton1]

z^2 - (7+i)z + 24 + 7i = 0
z= \frac{\ 7+i \pm \sqrt{(7+i)^2 - 4(24+7i)} }{ 2 }
z= \frac{\ 7+i \pm \sqrt{-47 - 14i }}{ 2 }
i stuck here

[Janitor]

I don't know whether you can simplify it further than that.

When I check the real part of what you have inside the root, I get 7^2 - 1 - 4*24, which is -48, not -47.

[newton1]

yes, it's -48
the answer in book is 3+4i , 4-3i

[Janitor]

I just substituted one of the book solutions into the original equation to check whether it is valid:

(3 + 4i) (3 + 4i) - (7 + i) (3 + 4i) + 24 + 7i = 0 + 0i.

So that book solution seems to be correct.

I also checked the other book solution, and it is also correct.

I do not know how you get from your form to the book form.

[Janitor]

To simplify sqrt(-48-14i), solve for real numbers a and b such that:

(a + bi)(a+bi)= -48-14i,

so that a^2 - b^2 = -48 and 2ab = -14.

You may be able to figure it out by doing that.

The next step could be that since a=-7/b,

(-7/b)^2 - b^2 = -48.

[newton1]

get it!!
Thank~~~

[HallsofIvy]

A standard way of handling \sqrt{-48-14i} is to convert to polar form and use DeMoivre's formula: this happens to have r= 50 and [theta]= 0.284 radians.
The square root will have r= \sqrt{50}= 5\sqrt{2} and theta= 0.142 radians or 3.283 radians. That is, the square roots are 5\sqrt{2}(cos(0.142)+ i sin(0.142))= 7+ i and 5\sqrt{2}(cos(3.283)+ i sin(3.283))= -7- i.