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 hoyy1kolko March 14th, 2011 08:45 PM

number and sets

1.Given z1=2-i and z2 =4+3i.Express

$\frac{1}{z1}+ \frac{6}{z2-z1}$ in the form a +bi ,where and b are real numbers.

My solution

$20+\frac{28i}{20}$

1+ 7/5i (is it correct?)

2.Solve the equation square root of (3x/(1-?x))=2

My solution

squaring the both sides

3x/1-?x=4
3x=4-4?x
9x^2+16z-6=0

 The Chaz March 14th, 2011 08:54 PM

Re: number and sets

$\sqrt{\frac{3x}{1 - \sqrt{x}}}= 2$

$\frac{3x}{1 - \sqrt{x}}= 4$
$3x= 4 - 4\sqrt{x}$
$3x - 4= -4\sqrt{x}$
$9x^2 -24x + 16= 16x$
$9x^2 -40x + 16= 0$
$(9x - 4)(x - 4)= 0$
...
x = 4 is out.

 MarkFL March 14th, 2011 09:03 PM

Re: number and sets

1.) $z_1=2-i$ and $z_2=4+3i$

$\frac{1}{z_1}+\frac{6}{z_2-z_1}$ I made the assumption that z_2-z_1 was the denominator of the 2nd term, you can use parentheses to make it clear. :mrgreen:

$\frac{1}{2-i}\cdot\frac{2+i}{2+i}+\frac{3}{1+2i}\cdot\frac{1-2i}{1-2i}=\frac{2+i}{5}+\frac{3-6i}{5}=\frac{5-5i}{5}=1-i$

 hoyy1kolko March 15th, 2011 07:41 AM

Re: number and sets

According to the solution of question 1 that you did,

how do you get 3/1+2i

from the part 6/z2-z1?

6/ 2+4i
3+3/2i (that's what I get ). Please tell me how you get 3/1+2i.

 The Chaz March 15th, 2011 07:44 AM

Re: number and sets

$\frac{6}{2 + 4i}= \frac{2\cdot 3}{2\cdot (1 + 2i)} = \frac{3}{1 + 2i}$

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