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 December 10th, 2009, 03:51 PM #1 Newbie   Joined: Dec 2008 From: Copenhagen, Denmark Posts: 29 Thanks: 0 Finding complex numbers that solves equation Hi all I'm trying to find the solution for the following equation $(z^2 + 2i)^2 + 2i= 0$ where z is a complex number. The solution should be written in the from $z= a + bi$ I've accomplished this some of the way $(z^2 + 2i)^2 + 2i= 0 \qquad \Longleftrightarrow \qquad (z^2)^2 + 4iz^2 + 4i^2 + 2i = 0 \qquad \Longleftrightarrow$ $z^2= -2i+\sqrt{-2i} \qquad$ or $\qquad z^2= -2i - \sqrt{-2i}$ Since $\sqrt{-2i}= \sqrt{2} - i\sqrt{2}$ the above can be written $z=\sqrt{\sqrt{2} - i(2+\sqrt{2})} \qquad$ or $\qquad z= -\sqrt{\sqrt{2} - i(2+\sqrt{2})} \qquad$ or $\qquad z= \sqrt{-\sqrt{2} -i(2-\sqrt{2})} \qquad$ or $\qquad z= -\sqrt{-\sqrt{2} -i(2 - \sqrt{2})}$ But I can't figure out how to express these as $z= a + bi$ Can anyone tell me how to do the last part?
 December 10th, 2009, 04:37 PM #2 Senior Member   Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0 Re: Finding complex numbers that solves equation Hi; Maybe this substitution would have been better: $\sqrt{-2 i}= 1 - i$
 December 11th, 2009, 03:55 PM #3 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 Re: Finding complex numbers that solves equation General approach - use polar form a + bi = r exp(iq), where r = ?(a^2 + b^2) and q=arctan(b/a), noting that q is multivalued. Use this form as an intermediate for taking square root, making sure you take into account multiplicity of the angle. After you have gotten the final results in polar, convert back to rectangle using a=rcosq and b=rsinq.
 December 13th, 2009, 06:51 AM #4 Newbie   Joined: Dec 2008 From: Copenhagen, Denmark Posts: 29 Thanks: 0 Re: Finding complex numbers that solves equation Thanks to both of you. Your ideas did help me understand the problem a little better, and made me look some more into my own textbooks. I actually ended up using another way to solve (I'm not sure what to call this in english, but directy translated here) "binomial equations": $(x+iy)^2= a + ib$ $x= \, +/- \sqrt{\frac{\sqrt{a^2 + b^2}+a}{2} \qquad$ $\qquad y= +/- \, \sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}$ $(z^2 + 2i)^2 + 2i= 0$ $z^2= -2i \, +/- \, \sqrt{-2i}$ Using the above method to find $\sqrt{-2i}$ gives me $\sqrt{-2i}= \sqrt{\frac{\sqrt{4+0} + 0}{2}} - i\sqrt{\frac{\sqrt{4+0}-0}{2}} = 1-i$ Then $z^2= 1 - 3i \qquad$ $\qquad z^2= -1-i$ $z_1= \sqrt{1-3i} \qquad$ , $\qquad z_2= - \, \sqrt{1-3i} \qquad$ , $\qquad z_3= \sqrt{-1-i} \qquad$ , $\qquad z_4= - \, \sqrt{-1-i}$ Using the method again gives me $z_1= \sqrt{\frac{\sqrt{10}+1}{2}} - i \sqrt{\frac{\sqrt{10}-1}{2}} \qquad$ , $\qquad z_2= -z_1$ $z_3= \sqrt{\frac{\sqrt{2}-1}{2}} - i \sqrt{\frac{\sqrt{2}+1}{2}} \qquad$ , $\qquad z_4= -z_3$ Which was actually the solution given in the course material.

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