If Z= [tan 2x -i(sin x+cos x)]/[1+i sin x] is completely imaginary, find x.(0<=x<=(pi/2) )
I understand that here we have to equate the real part to 0,but how do we isolate the real part.I tried multi plying both numerator and denominator by (1-2isinx),but it doesn't seem to be leading me any further.
"real"ize the denominator by multiplying top and bottom by 1-i sinx. Now you have
[tan 2x - i(sinx+cosx)][1 - i sinx]/[(1-i sinx)(1+i sinx)]
[tan2x - (sinx+cosx)(sinx) + i(junk)]/(1+sin^2 x)
to get zero real part, we need
tan2x = sin^2x + sinx cosx
any multiple of π clearly satisfies this equation.
It also makes Z have zero real part.
Thank you very much Steve!
To solve this problem, you need to isolate the real part of the given expression. Let's go through the steps together:
1. Start by multiplying the numerator and denominator by the complex conjugate of the denominator. In this case, the complex conjugate of (1 + i*sin(x)) is (1 - i*sin(x)). This is done to simplify the expression and eliminate the imaginary term in the denominator.
Z = [(tan(2x) - i(sin(x) + cos(x))) * (1 - i*sin(x))] / [(1 + i*sin(x)) * (1 - i*sin(x))]
2. Simplify the expression by using the distributive property and expanding the numerator:
Z = [tan(2x) - i*sin(x)*tan(2x) - i*cos(x) + i*cos(x)*sin(x)] / [1 - i^2*sin^2(x)]
3. Simplify the expression further:
Z = [tan(2x) - i*sin(x)*tan(2x) - i*cos(x) + i*cos(x)*sin(x)] / [1 + sin^2(x)]
4. Now, separate the real and imaginary parts of the expression:
Real part = tan(2x) - i*sin(x)*tan(2x) - i*cos(x) + i*cos(x)*sin(x)
Imaginary part = 0
5. Since the expression is completely imaginary (having no real part), we can set the real part equal to zero:
tan(2x) - i*sin(x)*tan(2x) - i*cos(x) + i*cos(x)*sin(x) = 0
Now, you can solve this equation to find the value of x within the given range (0 <= x <= pi/2). You can use algebraic methods or numerical methods, such as substitution or graphing, to find the solution.