Download Complex Number and Partial Fractions Problems with Solutions - Prof. John Elton and more Exams Mathematics in PDF only on Docsity! 1 Practice problems for test 1, part 1 Complex number problems. 1.Find the square roots of the complex number 1 3z i= − + . Solution: First let’s find the polar representation. r = |z| = magnitude of z = 2 2( 1) ( 3) 2− + = . By drawing a picture, you can see from your knowledge of special triangles that the angle the vector makes with the positive real axis is 120 degrees, or 2π/3 radians, that is, arg(z) = 2π/3 radians. Another way to do it is to use the equations x = r cos(θ) and y = r sin(θ) to solve for the argument θ. Since x = -1 and y = 3 , we get cos(θ) = -1/2, sin(θ) = 3 / 2 so θ = 2π/3 radians. One square root is gotten by taking half the angle and the square root of the magnitude, so 1 2(cos( / 3) sin( / 3)) 2((1/ 2) 3 / 2) ( 2 / 2) ( 6 / 2)w i i iπ π= + = + = + . The other square root is gotten by adding 2π/2 = π radians to the angle of this one, which is the same as just multiplying this one by –1, so the other one is 2 ( 2 / 2) ( 6 / 2)w i= − + − . 2. Find the cube roots of the complex number 1 3z i= − + . Solution: we already have the polar representation. One cube root is gotten by taking the cube root of the magnitude and one-third of the angle, so 1/3 1 2 (cos(2 / 9) sin(2 / 9))w iπ π= + . Now 2π/9 radians is 40 degrees, not a “special” angle, so we can just leave it like this, or if you prefer write a decimal approximation as w1 = .965 + i(.810), approximately. The other two cube roots are gotten from this by adding 2π/3 radians to the angle, and 2(2π/3) radians to the angle, so 1/3 1/32 2 2 2 8 8 2 9 3 9 3 9 92 (cos( ) sin( )) 2 (cos( ) sin( ))w i iπ π π π π π= + + + = + , and 1/3 1/32 4 2 4 14 14 3 9 3 9 3 9 92 (cos( ) sin( )) 2 (cos( ) sin( ))w i iπ π π π π π= + + + = + . I won’t bother with the decimals. The main thing to get out of this is that: one nth root of a complex number is gotten by taking the nth root of the magnitude and dividing the angle by n; the other n-1 roots are gotten from this one by adding multiples of 2π/n to it’s angle. The test question on complex numbers will be like this, only a different z, and a different n perhaps (like 4th root). Partial Fractions problems. These problems were taken from old tests: 1. 2 2 1 ( 1) x x dx x x − + +∫ Ans: ln(|x|) - tan -1(x) + C
