Complex Number Explanation?

Question

The two complex number Z1 = a/(1+i) and Z2 = 3/(1-2i) where a, b are real numbers. are such that Z1 + Z2 = 3. Calculate the value of a and b.

Okay, so i went about solving this question as an ordinary complex question.

a(1-i)/(1+i)(1-i) + b(1+2i)/(1-2i)(1+2i) = 3(1-i)(1+2i)

eventually getting,

(a-ai/2) + (b+2ib/5) = 9-3i

however, that doesn't seem to make the question any easier. and i'm not sure if multiplying the whole equation by 5 will solve the question. As it makes it just as complicated as it was in the beginning..so you guys, show me how to continue on or even better a quicker and easier method..if there is...thanks..ppl

Answer 1

I'm assuming that you meant Z2 = b/(1-2i)

The key is to equate the real parts and the imaginary parts. You also made a couple of errors.

Z1 + Z2 = 3
a/(1+i) + b/(1-2i) = 3
a(1-i)/[(1+i)(1-i)] + b(1+2i)/[(1+2i)(1-2i)] = 3

Note that you multiplied Z1 and Z2 by 1 so there is no need to multiply 3 by anything.

a(1-i)/2 + b(1+2i)/5 = 3

a/2 - ai/2 + b/5 + 2bi/5 = 3

(a/2 + b/5) + i(2b/5 - a/2) = 3

By matching the real and imaginary parts we get the equations

a/2 + b/5 = 3
2b/5 - a/2 = 0

Add these 2 equations

3b/5 = 3 ==> b=5

a/2 + 5/5 = 3 ==> a=4

So a=4 and b=5

Answer 2

You screwed up when you multiplied 3 by (1-i)*(1+2i); you multiplied a and b by 1 so they didn't change, but when you multiplied 3 by the conjugates, you destroyed the equality.

Better: 3 = a/(1+i) + b/(1-2i) =[a*(1-2i) + b*(1+i)]/(1+i)*(1-2i)
= [(a+b) +i*(-2a+b)]/(3-i) , or 9 - 3i = (a+b) +
i*(-2a+b) . Equating real and imaginary parts, we get a + b = 9 and -2a + b = -3. Thus, a = 4 and b = 5.

Answer 3

You have got the first step right, multiplying by the 1 - i and 1 + 2i.

However, you don't need to multiply the RHS by these numbers, because you are multiply both the top and bottom of each term by the same number.

So you should have (a - ai)/2 = (b + 2bi)/5 = 30

Multiply by 10 to eliminate the frax

5a - 5ai + 2b + 4bi = 30 = 30 + 0i

Then set the real and imaginary parts equal

5a + 2b = 30
(-5a + 4b) i = 0i

The i's cancel so

5a+ 2b = 30
-5a + 4b = 0

and solve for a and b

Answer 4

hi, permit's see. The argument for a complicated selection is given with the aid of the formulation arg(z)=arctan(b/a) , the place z=a+i*b subsequently, z= 8i, so a=0 and b=8, it relatively is undefined. The arctan for this "selection" is pi/2, with the aid of fact in case you inspect the tan functionality you will see that it is not defined at x=pi/2, and since the argument is defined interior the era ]-pi, pi], the fee is the the terrific option one.