Question #316162

Find square root of 1+i


1
Expert's answer
2022-03-23T15:25:43-0400

1+i=a+bi\sqrt{1+i}=a+bi

1+i=(a+bi)21+i=(a+bi)^2

1+i=(a2b2)+2abi1+i=(a^2-b^2)+2abi

Equating real and imaginary parts gives us;

1=a2b21=a^2-b^2

1=2ab

This is a simultaneous equation and so we are going to solve by substituting;

a=12ba=\frac{1}{2b}

1=(12b)2b2=14b2b21=(\frac{1}{2b})^2-b^2=\frac{1}{4b^2}-b^2

4b2=14b44b^2=1-4b^4

4b4+4b21=04b^4+4b^2-1=0

b2=4±424(4)(1)2(4)=1±22b^2=\frac{-4 \plusmn \sqrt{4^2-4(4)(-1)}}{2(4)}=\frac{-1 \plusmn \sqrt{2}}{2}

b2=122b^2=\frac{-1-\sqrt{2}}{2} has no real roots so we take the positive value

b=±2(21)2b=\plusmn \frac{\sqrt{2(\sqrt{2}-1)}}{2}

a=±221+2212a= \plusmn \frac{2\sqrt{\sqrt{2}-1}+\sqrt{2\sqrt{2}-1}}{2}

So finally the two possible roots of the complex number 1+i are ;

±(221+2212+2(21)2i)\plusmn (\frac{2\sqrt{\sqrt{2}-1}+\sqrt{2\sqrt{2}-1}}{2}+ \frac{\sqrt{2(\sqrt{2}-1)}}{2}i)


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