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 March 14th, 2011, 08:45 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 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 i just skip to the last two part $20+\frac{28i}{20}$ 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 Then i am stuck in here,please help.
 March 14th, 2011, 08:54 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 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.
 March 14th, 2011, 09:03 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs 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. $\frac{1}{2-i}\cdot\frac{2+i}{2+i}+\frac{3}{1+2i}\cdot\frac{1-2i}{1-2i}=\frac{2+i}{5}+\frac{3-6i}{5}=\frac{5-5i}{5}=1-i$
 March 15th, 2011, 07:41 AM #4 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 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.
 March 15th, 2011, 07:44 AM #5 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: number and sets $\frac{6}{2 + 4i}= \frac{2\cdot 3}{2\cdot (1 + 2i)} = \frac{3}{1 + 2i}$

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