Complex numbers questions

**swtdelicaterose** · Nov 26th 2009, 12:37 AM

So I missed the last week of class because I've been sick and with my textbooks stuck in my school lockers, I can't really get caught up on work. My friend was able to send me two sample questions they've been learning in class, but I have no idea what it means? Can someone please explain and teach me how to solve these questions?

Unsure of the notation too, the second z in the first question has a line over it, I don't think it's supposed to be z hat, but that's all I know.

1) Find the complex number z such that $(2+i)z + (4-i)\hat{z} = -1 + 5i$ .

2) If $u=-3 + 2i$ and $v = 3 + 5i$ are complex numbers, calculate the modulus of $\frac{u}{v}$ .

(Also, is modulus supposed to be like the programming modulus operation, or is it related? (i.e. 15 mod 10 = 5))

Thanks a lot, guys!

**tonio** · Nov 26th 2009, 01:54 AM

Originally Posted by swtdelicaterose

So I missed the last week of class because I've been sick and with my textbooks stuck in my school lockers, I can't really get caught up on work. My friend was able to send me two sample questions they've been learning in class, but I have no idea what it means? Can someone please explain and teach me how to solve these questions?

Unsure of the notation too, the second z in the first question has a line over it, I don't think it's supposed to be z hat, but that's all I know.

1) Find the complex number z such that $(2+i)z + (4-i)\hat{z} = -1 + 5i$ .

2) If $u=-3 + 2i$ and $v = 3 + 5i$ are complex numbers, calculate the modulus of $\frac{u}{v}$ .

(Also, is modulus supposed to be like the programming modulus operation, or is it related? (i.e. 15 mod 10 = 5))

Thanks a lot, guys!

Bad, bad thing to miss classes, specially classes dealing with maths...tsk,tsk,tsk.

Anyway, what you have here requires you first read and understand the basics of complex numbers (conjugate, modulus, operations, etc.), so that you'll "have some idea" what's this about. otherwise it's like asking someone to solve your homework for you and to explain you all this in this forum would be too long.

So study the basics (there are tens of thousands of internet sites with explanations on this: jsut google "complex numbers" and surf the sites) , and then, if you're still stuck somewhere, write back and ask.

Tonio

**Shanks** · Nov 26th 2009, 01:57 AM

Have you try your best to solve it?
where are you stuck on?

**stapel** · Nov 26th 2009, 04:35 AM

To learn some of the basics of complex numbers, try here.

Originally Posted by swtdelicaterose

1) Find the complex number z such that $(2+i)z + (4-i)\hat{z} = -1 + 5i$ .

Write z as a + bi. Then multiply out the left-hand side, collecting like terms and simplifying, so you get a complex number in terms of "a" and "b". Set the real and imaginary coefficients equal, and solve the resulting system of equations for the values of "a" and "b".

Originally Posted by swtdelicaterose

2) If $u=-3 + 2i$ and $v = 3 + 5i$ are complex numbers, calculate the modulus of $\frac{u}{v}$ .

The "modulus" is the "size", found by squaring the real and imaginary coefficients, summing the squares, and taking the square root of the result. So you'll need to rationalize the denominator (the lesson at the link explains how), and then do the modulus process.

After you've studied the lesson, please attempt these exercises. If you get stuck, please reply showing your work so far, so we can "see" where you're still having difficulty. Thank you!

**swtdelicaterose** · Nov 27th 2009, 12:46 AM

Okay I think I solved number 2:

$ \frac{-3+2i}{3+5i} \times \frac{3-5i}{3-5i} $

$ = \frac{9+15i+6i-10i^2}{9-25i^2} = \frac{21i + 1}{34}$

modulus is then:
$ \sqrt{\frac{21^2 + 1^2}{1156}} = \sqrt{\frac{442}{1156}} $
Thanks a lot, still working on first question.

Edit: Still can't figure out what $\bar{Z}$ is. I know it's the conjugate of a complex number, but is there a property I use to cancel out the two different Z's?

**swtdelicaterose** · Nov 27th 2009, 01:28 AM

1) Find the complex number z such that $(2+i)z + (4-i)\hat{z} = -1 + 5i$ .
(failed attempt)

$ 2z + iz + 4\bar{z} - i\bar{z} = -1 + 5i $
iz and -iz cancel out

$ 6z = -1 + 5i$

$ z = \frac{-1}{6} + \frac{5}{6}i $

But it's wrong, so I guess z != z conjugate.

What else can I use?

**stapel** · Nov 27th 2009, 04:06 AM

Try using the hint suggested earlier. Since $z$ is some complex number, then $z\, =\, a\, +\, bi$ for some real numbers $a$ and $b]$ .

Now multiply out and equate.

**swtdelicaterose** · Nov 27th 2009, 10:10 AM

Oh, I think I kinda get it:

$(2+i)(a+bi)+(4-i)(a-bi)=-1-5i$

$2a + 2bi + ia + b(i^2) + 4a - 4bi -ia + b(i^2) = -1-5i$

Simplifying I get:

$4a - 4bi = -1 -5i$

$a - bi = -\frac{1}{4} - \frac{5}{4}i$

I'm not sure about the last step, because I don't know how to convert the conjugate into a normal complex. My textbook doesn't list any property for that. Or does setting bi to a positive just change the sign of the i on the right side?

$\bar{z_1}\bar{+}\bar{z_2} = \bar{z_1}+\bar{z_2}$

So If I do the conjugate of a-bi, I should get a+bi:

$a+bi = -\frac{1}{4} + \frac{5}{4}i$

and since 1/4 is a real number, the conjugate remains the same. Since the second par is a number in terms of i, the conjugate is a sign change.

I don't think this is the correct answer though. Anyone know what I did wrong?

**swtdelicaterose** · Nov 27th 2009, 04:57 PM

Looks like it should be:

$6a-2bi = -1+5i$

From here, I don't know how to get it in terms of z since the real and imaginary coefficients are different. How can I make them the same?

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