Calculus

posted by .

In these complex exponential problems, solve for x:

1)e^(i*pi) + 2e^(i*pi/4)=?

2)3+3=3i*sqrt(3)=xe^(i*pi/3)


MY attempt:

I'm not really sure of what they are asking.
For the 1st one I used the e^ix=cos(x)+i*sin(x)
and got -1+sqrt(2) +sqrt(2)i

2) I solved for x and fot (3+3i*sqrt(3))/(1/2+i*sqrt(3)/2)

  • Calculus -

    your 1st answer is correct

    #2.
    x(1/2 + √3/2 i) = 3+3i√3
    x = (3+3√3 i))/[1/2 (1+√3 i)]

    now rationalize by multiplying by conjugate

    x = 3(1+√3 i)* 2(1-√3 i)/(1-3)
    x = -3(1+√3 i))(1-√3 i)
    x = -3(1-3)
    x = 6

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Math Help please!!

    Could someone show me how to solve these problems step by step.... I am confused on how to fully break this down to simpliest terms sqrt 3 * sqrt 15= sqrt 6 * sqrt 8 = sqrt 20 * sqrt 5 = since both terms are sqrt , you can combine …
  2. Calculus - Second Order Differential Equations

    Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'(0)=4, …
  3. Calculus - Second Order Differential Equations

    Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 …
  4. Calculus

    Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …
  5. Math/Calculus

    Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else?
  6. Precalculs

    I have no idea how to do these type of problems. -------Problem-------- Solve each equation on the interval 0 less than or equal to theta less than 2 pi 42. SQRT(3) sin theta + cos theta = 1 ---------------------- There is an example …
  7. TRIG

    what is the exact value of sin^2 pi/6-2 sin pi/6 cos pi/6 +cos^2 (-pi/6)?
  8. calculus

    Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= …
  9. maths

    Choose the option that gives the solution of the initial-value problem dy dx=(1 + 2 cos2x^2)/y (y > 0), y= 1 when x = 0. Options A y = 1+ 2sin^2 x B y = (1 + 2 sin x)^2 C y = (4x + cos(2x))^2 D y = 4x + cos^2 x E y =sqrt(4x + cos(2x) …
  10. Calculus AP

    hi again im really need help TextBook: James Stewart:Essential Calculus, page 311. Here the problem #27: First make a substitution and then use integration by parts to evaluate the integral. Integral from sqrt(pi/2) TO sqrt(pi)of θ^3 …

More Similar Questions