Question

Solve for A, B, C, ..., I in:

sqrt(-2) = +jA and -jA;
8 + sqrt(-5) = B + jC and B - jC;
(2 - 7j) + (6 + 8j) = D + jE;
j^35 = F + jG

Answer 1

To solve the given equations, we need to isolate the variables A, B, C, ..., I in each equation. By performing the necessary operations, we can find the values of these variables in each equation.

Step-by-step explanation:

To solve for A, B, C, ..., I in the given equations, we need to perform the necessary operations to isolate the variables. Let's go step by step for each equation:

Equation 1: sqrt(-2) = +jA and -jA

To solve for A, we can square both sides of the equation: -2 = (jA)^2 = -A^2. Taking the square root of -2 would give us the imaginary number √-2 = +j√2 and -j√2. So A = √2 or A = -√2.

Equation 2: 8 + sqrt(-5) = B + jC and B - jC

Similarly, we can solve for B and C. By equating real parts and imaginary parts separately, we get B = 8 and C = √5 or C = -√5.

Equation 3: (2 - 7j) + (6 + 8j) = D + jE

Add the real parts and imaginary parts separately: 2 + 6 = D and -7 + 8 = E. Therefore, D = 8 and E = 1.

Equation 4: j^35 = F + jG

Since j is a special complex number used to represent imaginary numbers, j^35 = j^(4*8 + 3) = j^3 = -j. Therefore, F = 0 and G = -1.