What is the reciprocal of #6+i#?

Question

the answer is #(6-i)/37# but i dont know how. please explain!

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Answer 1 · 2018-06-01T21:18:33

#(6-i)/(37)#

#6+i#

reciprocal:

#1/(6+i)#

Then you have to multiply by the complex conjugate to get the imaginary numbers out of the denominator:

complex conjugate is #6+i# with the sign changed over itself:

#(6-i)/(6-i)#

#1/(6+i)*(6-i)/(6-i)#

#(6-i)/(36+6i-6i-i^2)#

#(6-i)/(36-(sqrt(-1))^2)#

#(6-i)/(36-(-1))#

#(6-i)/(37)#

Answer 2 · 2018-06-01T21:18:51

The reciprocal of #a# is #1/a#, therefore, the reciprocal of #6+i# is:

#1/(6+i)#

However, it is bad practice to leave a complex number in the denominator.

To make the complex number become a real number we multiply by 1 in the form of #(6-i)/(6-i)#.

#1/(6+i)(6-i)/(6-i)#

Please observe that we have done nothing to change the value because we are multiplying by a form that is equal to 1.

You may be asking yourself; "Why did I choose #6-i#?".

The answer is because I know that, when I multiply #(a+bi)(a-bi)#, I obtain a real number that is equal to #a^2+b^2#.

In this case #a = 6# and #b=1#, therefore, #6^2+1^2 = 37#:

#(6-i)/37#

Also, #a+bi# and #a-bi# have special names that are called complex conjugates.