Find complex number z from equation

[prehisto]

Homework Statement

Find complex number z from equation
Sinz+cosz=4

Homework Equations

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

 

[haruspex]

\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 get something a little different. Check your signs. If you still get the same answer, please post your working.

Next step could be to use Quadratic formula,but the roots are complicated

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

 

[prehisto]

I get something a little different. Check your signs. If you still get the same answer, please post your working.


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

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

so
t_1,2=-8i+(-)\sqrt{(8i)^2-4(1+i)(1-i)}/(2(1+i))

t_1,2=-8i+(-)\sqrt{-64-8}/(2(1+i))

I ment complicated roots .
I cat simplify them further.

 

[haruspex]

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

You've changed two signs. If you'd only changed the first one we'd now agree. Please check again.

so
t_1,2=-8i+(-)\sqrt{(8i)^2-4(1+i)(1-i)}/(2(1+i))

t_1,2=-8i+(-)\sqrt{-64-8}/(2(1+i))

I can't simplify them further.

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^iθ form because you still have to find z.

 

[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.

 

[haruspex]

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.

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