# Thread: complex numbers question

1. ## complex numbers question

hi guys, this is my first post.

wondered is you could help me with these two questions, its the methods i need more than the answers

1) the complex numbers z and w satisfy the simultaneous equations

2z + iw = -1
z - w = 3 + 3i

a) use algebra to find z, giving your answer in the form a+ib, where a and b are real

b) carculate arg z, giving your answer in radians to 2dp.

2) the complex numbers z and w are given by

z = A / (1-i)
w = B / (1-3i)

where A and B are real numbers. given that z + w = i

a) find the value of A and the value of B

b) for these values of A and B, find tan[arg(w-z)]

2. Originally Posted by djr8793
hi guys, this is my first post.

wondered is you could help me with these two questions, its the methods i need more than the answers

1) the complex numbers z and w satisfy the simultaneous equations

2z + iw = -1
z - w = 3 + 3i

a) use algebra to find z, giving your answer in the form a+ib, where a and b are real
1. $2z + iw = -1$
$z - w = 3 + 3i$.

From equation 2 we can see that $w = z - 3 - 3i$.

Substituting into equation 1 gives

$2z + i(z - 3 - 3i) = -1$

$2z + iz - 3i + 3 = -1$

$(2 + i)z = -4 + 3i$

$z = \frac{-4 + 3i}{2 + i}$

$z = \frac{(-4 + 3i)(2 - i)}{(2 + i)(2 - i)}$

$z = \frac{-8 + 4i + 6i + 3}{4 + 1}$

$z = \frac{-5 + 10i}{5}$

$z = -1 + 2i$.

To find $\arg{z}$, notice that $\arg{z}$ is in the second quadrant.

So work out $\theta$ for the same complex number in the first quadrant and then use symmetry:

$\theta = \arctan{\frac{2}{1}}$

$= \arctan{2}$.

So this means that $\arg{z} = \pi - \arctan{2}$.

3. Originally Posted by djr8793
2) the complex numbers z and w are given by

z = A / (1-i)
w = B / (1-3i)

where A and B are real numbers. given that z + w = i

a) find the value of A and the value of B

b) for these values of A and B, find tan[arg(w-z)]
$z + w = i$, so

$\frac{A}{1 - i} + \frac{B}{1 - 3i} = i$

$\frac{A(1 + i)}{(1 - i)(1 + i)} + \frac{B(1 + 3i)}{(1 - 3i)(1 + 3i)} = 0 + i$

$\frac{A + Ai}{2} + \frac{B + 3Bi}{10} = 0 + i$

$\frac{5A + 5Ai + B + 3Bi}{10} = 0 + i$

$5A + B + 5Ai + 3Bi = 10(0 + i)$

$5A + B + (5A + 3B)i = 0 + 10i$.

Now equate real and complex coefficients:

$5A + B = 0$ and $5A + 3B = 10$.

From equation 1 we can see $B = -5A$.

Substituting into equation 2 gives:

$5A + 3(-5A) = 10$

$5A - 15A = 10$

$-10A = 10$

$A = -1$.

And since $B = -5A$

$B = -5(-1)$

$B = 5$.

b) $z - w = -\frac{1}{1 - i} - \frac{5}{1 - 3i}$

$= -\frac{1(1 + i)}{(1 - i)(1 + i)} - \frac{5(1 + 3i)}{(1 - 3i)(1 + 3i)}$

$= \frac{-1 - i}{2} + \frac{-5 - 15i}{10}$

$= \frac{-5 - 5i - 5 - 15i}{10}$

$= \frac{-10 - 20i}{10}$

$= -1 - 2i$.

Working out angle $\theta$ in the first quadrant gives:

$\theta = \arctan{\frac{2}{1}}$

$\theta = \arctan{2}$.

And since $\arg{(z - w)}$ will be in the third quadrant, and tangent is positive in the third quadrant, this means

$\tan{\arg{(z - w)}} = \tan{\theta}$

$= \tan{\arctan{2}}$

$= 2$.

4. thanks prove it, i understannd it all now