Αποτελέσματα Αναζήτησης
The problems covered include finding the general value of log(1+i)+log(1-i), evaluating logarithms of expressions involving trigonometric functions, and using properties of logarithms to simplify complex logarithmic expressions.
Complex numbers - Exercises with detailed solutions 1. Compute real and imaginary part of z = i¡4 2i¡3: 2. Compute the absolute value and the conjugate of z = (1+ i)6; w = i17: 3. Write in the \algebraic" form (a+ib) the following complex numbers z = i5 +i+1; w = (3+3i)8: 4. Write in the \trigonometric" form (‰(cosµ +isinµ)) the following ...
Examples for Complex numbers Question (01) (i) Find the real values of x and y such that (1 ) 2 (2 3 ) 3 3 i x i i y i i i i − + + + + =− − + (ii) Find the real values of x and y are the complex numbers 3−ix y2 and − − −x y i2 4 conjugate of each other. (iii) Find the square roots of 4 4+i (iv) Find the complex number Z satisfying ...
16. A bacteria culture starts with 10 00 bacteria and the number doubles every 40 minutes. (a) Find a formula for the number of bacteria at time t. (b) Find the number of bacteria after one hour. (c) After how many minutes will there be 50 000 bacteria?
30 Απρ 2024 · Complex Logarithm. The complex logarithm is an extension of the concept of logarithmic functions involving complex numbers (represented by log z). Mathematically, written as. log (z) = log (r ⋅ e iθ) = ln (r) + i (θ + 2nℼ) Here, z = r ⋅ e iθ = the complex number. r = |z| = the absolute value of z.
Complex logarithm Complex power function Definition Principal value of ln(|z|) Properties of the logarithm You have to be careful when you use identities like ln(z 1z 2) = ln(z 1)+ln(z 2), or ln z 1 z 2 =ln(z 1)−ln(z 2). They are only true up to multiples of 2πi. For instance, if z 1 = i = exp(iπ/2) and z 2 = −1=exp(iπ), ln(z 1)=i π 2 ...
Main examples: I = [0; 2 ) or I = ( ; ] : arg[0;2 )( 3 ; i) = 2. arg( ; ]( = argI(z) has a cut line where the value jumps by 2 . Complex logarithms. To solve ew = z , for z 6= 0 : Let w = u + iv ; so ew = eueiv ; and write z = jzj e i : eueiv = jzj e i , eu = jzj ; eiv = ei : Infinitely many solutions w : u = log jzj ; v = + 2 k.
