# Find complex number z from equation

1. Dec 9, 2013

### prehisto

1. The problem statement, all variables and given/known data
Find complex number z from equation
Sinz+cosz=4

2. Relevant equations

3. The attempt at a solution
I rewrite sinz+cosz=4 in form:
$\frac{(exp[i*z]-exp[-i*z])}{2i}$+$\frac{(exp[i*z]+exp[-i*z])}{2}$=4
then subsitute t=exp[i*z]

So I gain t2(1+i)+8ti-1+i=0,I think i can rewrite it in form t2a+tb-c=0
Next step could be to use Quadratic formula,but the roots are complicated, maybe I have mistaken ?

so

Last edited by a moderator: Dec 9, 2013
2. Dec 9, 2013

### haruspex

Do you mean complex? Why would that be wrong? Post whatever you get.

3. Dec 10, 2013

### prehisto

Yes,i got something different now.
t2(1+i)-8ti+(1-i)=0

so
t1,2=-8i+(-)$\sqrt{(8i)^2-4(1+i)(1-i)}$/(2(1+i))

t1,2=-8i+(-)$\sqrt{-64-8}$/(2(1+i))

I ment complicated roots .
I cat simplify them further.

4. Dec 10, 2013

### haruspex

You've changed two signs. If you'd only changed the first one we'd now agree. Please check again.
Well, you can simplify -64-8, and get rid of the complex denominator. I see no difficulty in reducing it to the usual a+ib form. But actually you will want it in re form because you still have to find z.

5. Dec 10, 2013

### prehisto

so finally i have t^2(1+i)-8ti+i-1=0

Unfortunely i do not see how to reduce it. How do you recommend to get rid of complex denominator.

6. Dec 10, 2013

### haruspex

That's a standard procedure: multiply top and bottom by its conjugate.