prehisto said:[itex]\frac{(exp[i*z]-exp[-i*z])}{2i}[/itex]+[itex]\frac{(exp[i*z]+exp[-i*z])}{2}[/itex]=4
then subsitute t=exp[i*z]
So I gain t2(1+i)+8ti-1+i=0
Do you mean complex? Why would that be wrong? Post whatever you get.Next step could be to use Quadratic formula,but the roots are complicated
haruspex said: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.
You've changed two signs. If you'd only changed the first one we'd now agree. Please check again.prehisto said:Yes,i got something different now.
t2(1+i)-8ti+(1-i)=0
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 reiθ form because you still have to find z.so
t1,2=-8i+(-)[itex]\sqrt{(8i)^2-4(1+i)(1-i)}[/itex]/(2(1+i))
t1,2=-8i+(-)[itex]\sqrt{-64-8}[/itex]/(2(1+i))
I can't simplify them further.
That's a standard procedure: multiply top and bottom by its conjugate.prehisto said: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.
To find the complex number z from an equation, we need to first set the equation equal to z. Then, we can use algebraic manipulation to isolate the complex number z on one side of the equation. Once we have z by itself, we can determine the real and imaginary parts of z by comparing the equation to the standard form for complex numbers, a + bi.
Yes, the quadratic formula can be used to find the complex number z. However, the quadratic formula will only give us the two solutions for z, and we need to verify which one is correct by plugging it back into the original equation. It is also important to remember that the discriminant of the quadratic equation (b^2 - 4ac) must be negative in order for the solutions to be complex numbers.
Finding the complex number z from an equation can be thought of as finding the point on the complex plane that satisfies the given equation. The real part of z corresponds to the x-coordinate on the complex plane, while the imaginary part corresponds to the y-coordinate. So, graphing the equation and finding the intersection with the x-axis will give us the real part of z, and the intersection with the y-axis will give us the imaginary part of z.
Yes, complex numbers can be represented in polar form. In polar form, a complex number is written as r(cosθ + isinθ), where r is the magnitude of the complex number and θ is the angle between the real axis and the vector representing the complex number on the complex plane. This form is useful for performing operations like multiplication and division on complex numbers.
When finding the complex number z from an equation, we can use the property of complex conjugates to simplify the process. The complex conjugate of a+bi is a-bi, and multiplying a complex number by its conjugate results in a real number. So, if we have a complex number in the form a+bi in our equation, we can multiply both sides by its conjugate to eliminate the imaginary part and solve for the real part of z.