Write the expression as a complex number in standard form?

#1/(3-5i)-(6-2i)#

1 Answer
Shwetank Mauria
Nov 21, 2017

#1/(3-5i)-(6-2i)=-201/34+73/34i#

Explanation:

#1/(3-5i)-(6-2i)#

= #1/(3-5i)xx(3+5i)/(3+5i)-(6-2i)#

= #(3+5i)/(3^2+5^2)-6+2i#

= #(3+5i)/34-6+2i#

= #3/34+5/34i-6+2i#

= #(3/34-6)+(5/34+2)i#

= #(3-204)/34+(5+68)/34i#

= #-201/34+73/34i#

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