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 May 5th, 2011, 10:32 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Complex number 1. Determine the value of m if z= ?5+mi/4-2?5i is a pure complex number and find this real number. 2. Express (1+i)^3/(1-i)^5 in the form a+ib, where a and b are real. Hence, find the argument and modulus of this complex number. I need help. I can't even solve one of the question. It is an exam question. Please show me step by step how you solving the question. Thank you.
 May 5th, 2011, 10:46 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Complex number 1.) What is a pure complex number? 2.) One way to do this is through use of Euler's formula: $\frac{(1+i)^3}{(1-i)^5}=\frac{$$\sqrt{2}\(e^{\frac{\pi}{4}i}$$\)^3}{ $$\sqrt{2}\(e^{-\frac{\pi}{4}i}$$\)^5}=\frac{1}{2}e^{2\pi i}=\frac{1}{2}\cdot1=\frac{1}{2}$ Now you should be able to easily compute the argument and modulus.
 May 5th, 2011, 10:55 PM #3 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Complex number MarkFL, I think by 'pure complex' is meant the same as 'pure imaginary', i.e. it lies on the imaginary axis. -Ormkärr-
 May 5th, 2011, 11:08 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Complex number I thought that was probably the case, but I wasn't sure...thanks! In that case, it appears we should set: $\sqrt{5}+\frac{m}{4}i=0$ $\sqrt{5}=-\frac{m}{4}i$ $m=-\frac{4\sqrt{5}}{i}=(i^2)\frac{4\sqrt{5}}{i}=4\sqr t{5}i$ I don't know what you mean by "find this real number" because z and m are imaginary.
 May 5th, 2011, 11:30 PM #5 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Complex number MarkFL, You are being deceived by the poor type-setting employed in the OP. I am convinced it should read $z\=\ \frac{\sqrt{5}+mi}{4-2\sqrt{5}i}.$ In this case, I think the answer is real. -Ormkärr-
 May 5th, 2011, 11:39 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Complex number Aw jeeez! This isn't the first time such a thing has happened! I'm so entrenched in the use of parentheses for clarity, I tend to assume everyone else is too.
 May 5th, 2011, 11:44 PM #7 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Complex number It's probably better for your health to give the benefit of the doubt as you do! I for one should be scoping out good deals on cemetery plots, given my level of faith in my fellow man. -Ormkärr-
 May 6th, 2011, 04:22 AM #8 Global Moderator   Joined: Dec 2006 Posts: 18,245 Thanks: 1439 2.  $(1\,+\,i)^3/(1\,-\,i)^5\,=\,(1\,+\,i)^8/32\,=\,(2i)^4/32\,=\,1/2$
 May 6th, 2011, 08:44 AM #9 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Complex number I still don't understand question 1.
 May 6th, 2011, 08:55 AM #10 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Complex number hoyy1kolko, If we write $z\=\ \frac{\sqrt{5}+mi}{4-2\sqrt{5}i},$ then we want to "realize" (my version of "rationalize", not standard) the denominator, so multiply top and bottom by the complex conjugate of the denominator, namely 4+2?5i to get \begin{align*} z\ &=\ \left(\frac{\sqrt{5}+mi}{4-2\sqrt{5}i}\right)\left(\frac{4+2\sqrt{5}i}{4+2\sq rt{5}i}\right)\\ &=\ \frac{(4\sqrt{5}-2m\sqrt{5})+(4m+10)i}{36}. \end{align*} At this point, since you want a 'pure imaginary' number, you must have the real part of the above vanish, so you set $4\sqrt{5}-2m\sqrt{5}\ =\ 0 \quad \Rightarrow\\ m\ =\ 2,$ and you are done. -Ormkärr-

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