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- SOLVE THE INTEGRAL DOUBLE SHOWN IN THE PICTURE. (REMEMBER THAT YOU CAN MAKE THE CHANGE FROM COORDINATES TO POLAR COORDINATES)Find the product z,z2 and the quotient 2. Express your answers in polar form. z1 = 3(cos(250°) + i sin(250°)), z2 = 25(cos(200°) + i sin(200°)) Z24. Evaluate: sin (1+ 2j). Express in polar form
- Write z1 and Z2 in polar form. (Express 0 in radians. Let 0 <0 < 2x.) Z1 = 2/3 - 2i, z2 : -1 + i Z1 = Z2 Z1 and (Express your answers in polar form with 0 expressed in radians.) Find the product z1z2 and the quotients Z1 Z2 ZĄZ2 = Z1 22 Z1 ||Choose the correct region of integration 5 √³x y dy dx. Assume that in each figure, the horizontal axis is the x-axis and the vertical axis is the y-axis.Find the product z1z2 and the quotient . Express your answers in polar form. Z2 Z1 = V 2( cos 4 +i sin( 5), Z2 8(cos(n) + i sin(x)) 4 ZĄZ2 = Z1 Z2
- Write z1 and z2 in polar form. (Express 0 in radians. Let 0 < 0 < 27.) z1 = 4/3 - 4i, z2 = -1 + i Z1 = %D Z2 %3D Z1 Find the product z1z2 and the quotients and (Express your answers in polar form with 0 expressed in radians.) 22 Z1 ZĄZ2 = Z1 Z2 Z1 IIChange the rectangular coordinate (-2,2/3) to the polar coordinate. Give one 4. answer with a positive r and a negative r with 0<0<2n .3-Use polar coordinate to evaluate f, S a2-x2 dxdy a2-x2 ala