# Complex equation

1. Apr 6, 2004

### 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

2. Apr 6, 2004

### 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.

Last edited: Apr 6, 2004
3. Apr 6, 2004

### newton1

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

4. Apr 6, 2004

### 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.

Last edited: Apr 6, 2004
5. Apr 6, 2004

### 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.

6. Apr 6, 2004

get it!!
Thank~~~

7. Apr 6, 2004

### HallsofIvy

Staff Emeritus
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.