ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ziaxadon on November 29, 2017, 12:13:38 am
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Hey guys, I am having trouble doing the following problems.
Use the Cartesian method to find the square roots of each of the following.
a) 1+i
b) i
Thanks in advance!!
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If we square a complex number, we get another complex number. To do this sort of question using a Cartesian method, therefore, start with:
^2=1+i\\<br />a^2+2abi-b^2=1+i\\<br />a^2-b^2=1, 2abi=i)
Then solve simultaneously.
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Thanks so much for your help. I used your method to solve simultaneously and got a surd as an answer however my textbook saids that this surd could be further simplified and I don't know how they achieved this. I checked the textbook's answer on my CAS and both the CAS and the textbook displayed the same answer. Can anyone please tell me what simplification techniques are used and how you can obtain the simplest form to both "a" and "b". Attached is my worked solution and the steps that I am having trouble with.
Thanks in advance!!!
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I can't see your working, but I have continued mine below.
^2-b^2=1\\<br />\frac{1}{4b^2}-b^2=1\\<br />\frac{1}{4b^2}-\frac{4b^4}{4b^2}=1\\<br />\frac{1-4b^4}{4b^2}=1\\<br />1-4b^4=4b^2\\<br />0=4b^4+4b^2-1\\<br />0=\left[\left(2b^2\right)^2+2\times 2b\times 1+1\right]-1-1\\<br />0=\left(2b^2+1\right)^2-2\\<br />2=\left(2b^2+1\right)^2\\<br />2b^2+1=\pm\sqrt{2}\\<br />b^2=\frac{-1\pm\sqrt{2}}{2}\\<br />b=\pm\sqrt{\frac{-1\pm\sqrt{2}}{2}}\\<br />b=\pm\sqrt{\frac{-1\pm\sqrt{2}}{2}}\times\frac{\sqrt{2}}{\sqrt{2}}\\<br />b=\pm\sqrt{\frac{2\left(-1\pm\sqrt{2}\right)}{2^2}}\\<br />b=\frac{\pm\sqrt{2\left(-1\pm\sqrt{2}\right)}}{2}\\<br />\frac{\pm\sqrt{2\left(-1-\sqrt{2}\right)}}{2}\notin\mathbb{R}\\<br />\therefore b=\frac{\pm\sqrt{2\left(-1+\sqrt{2}\right)}}{2}\\<br />\therefore a=\frac{1}{2\left(\frac{\pm\sqrt{2\left(-1+\sqrt{2}\right)}}{2}\right)}\\<br />a=\frac{1}{\pm\sqrt{2\left(-1+\sqrt{2}\right)}}\\<br />a=\frac{1}{\pm\sqrt{2\left(-1+\sqrt{2}\right)}}\times\frac{\sqrt{\left(1+\sqrt{2}\right)}}{\sqrt{\left(1+\sqrt{2}\right)}}\\<br />a=\frac{\sqrt{\left(1+\sqrt{2}\right)}}{\pm\sqrt{2\left(-1+\sqrt{2}\right)\left(1+\sqrt{2}\right)}}\\<br />a=\frac{\sqrt{\left(1+\sqrt{2}\right)}}{\pm\sqrt{2\left(-1+2\right)}}\\<br />a=\pm\frac{\sqrt{\left(1+\sqrt{2}\right)}}{\sqrt{2}}\\<br />a=\pm\frac{\sqrt{\left(1+\sqrt{2}\right)}}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}\\<br />a=\pm\frac{\sqrt{2\left(1+\sqrt{2}\right)}}{2}\\<br />\left(a,b\right)=\left(\pm\frac{\sqrt{2\left(1+\sqrt{2}\right)}}{2},\frac{\pm\sqrt{2\left(-1+\sqrt{2}\right)}}{2}\right)\\<br />\therefore\sqrt{1+i}=\frac{\sqrt{2\left(1+\sqrt{2}\right)}}{2}+\frac{\sqrt{2\left(-1+\sqrt{2}\right)}}{2}i\\<br />\text{or }\sqrt{1+i}=-\frac{\sqrt{2\left(1+\sqrt{2}\right)}}{2}-\frac{\sqrt{2\left(-1+\sqrt{2}\right)}}{2}i<br />)
Note that some publications will define a "principal square root", giving only one answer, however, this is not necessary in the Specialist course