Who knows the square root of i
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Square Root of i

Do you remember the imaginary number i, which stands for −1−−−√ ?  We couldn't describe −1−−−√ with a real number, since the square of a positive number is positive, and the square of a negative number is positive.  There is no real number whose square is negative.  The square root of a real number is not always a real number.  It turns out that −1−−−√ is a rather curious number, which you can read about in Imaginary Numbers.

But have you ever thought about i√ ?  Won't we need a j, or some other invention to describe it?  Actually, no.  It turns out that i√ is another complex number.  We won't need a j.

When we first encountered the number i, we also learned about complex numbers, or numbers of the form a+bi.  It turns out that square roots of complex numbers are always other complex numbers.

Consider   2+3i   for the moment.  We can square that number:

(2+3i)2=(2+3i)(2+3i)=4+6i+6i+9i2=4+12i−9=−5+12i

Therefore, the square root of   −5+12i   is   2+3i.   So now we have demonstrated one case where the square root of a complex number is another complex number.

An algebraic derivation

So let's assume that there is a number   a+bi   which represents i√.  Since

(a+bi)2=a2+2abi+b2i2=(a2−b2)+(2ab)i

and since this result should equal the number i, we obtain the following system of equations:

{a2−b22ab=0=1

When the first equation is solved for the variable a, we learn that the variables a and b are either equal or opposites.  From the second equation, we know that the product is a positive one-half, so the variables a and b must be equal.  (If they were opposite, the product would have been negative.)  Therefore, the second equation becomes   2a2=1,   whose solution is   a=±2–√2.

Therefore, the number i has two square roots (just like positive numbers do).  They are   2–√2+2–√2i   and   −2–√2−2–√2i.

Step-by-step explanation:

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~Feitan Portor~

78.59 is the mean

step-by-step explanation: