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March 14th, 2011, 07:45 PM  #1 
Senior Member Joined: Dec 2010 Posts: 233 Thanks: 0  number and sets
1.Given z1=2i and z2 =4+3i.Express in the form a +bi ,where and b are real numbers. My solution i just skip to the last two part 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=44?x 9x^2+16z6=0 Then i am stuck in here,please help. 
March 14th, 2011, 07:54 PM  #2 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: number and sets ... x = 4 is out. 
March 14th, 2011, 08:03 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs  Re: number and sets
1.) and I made the assumption that z_2z_1 was the denominator of the 2nd term, you can use parentheses to make it clear. 
March 15th, 2011, 06:41 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/z2z1? 6/ 2+4i 3+3/2i (that's what I get ). Please tell me how you get 3/1+2i. 
March 15th, 2011, 06:44 AM  #5 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: number and sets 

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