Question 147558
{{{(4)/(2+5i)}}} Start with the given expression.



{{{((4)/(2+5i))((2-5i)/(2-5i))}}} Multiply both numerator and denominator by the complex conjugate of {{{2+5i}}}



{{{((4)(2-5i))/((2+5i)(2-5i))}}} Combine the fractions



{{{((4)(2-5i))/(29)}}} Foil the denominator to get {{{(2+5i)(2-5i)=4-10i+10i-25i^2=4-25(-1)=29}}}



{{{(8-20i)/(29)}}} Distribute



So {{{(4)/(2+5i)=(8-20i)/(29)}}}



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{{{i^(-35)}}} Start with the given expression.



{{{1/i^(35)}}} Flip the fraction to make the exponent positive.


Now let's find evaluate {{{i^(35)}}}:


First take the exponent 35 and divide by 4.
When it leaves a remainder of 0, the answer is {{{1}}}.
When it leaves a remainder of 1, the answer is {{{sqrt(-1)=i}}}.
When it leaves a remainder of 2, the answer is {{{-1}}}.
When it leaves a remainder of 3, the answer is {{{-sqrt(-1)=-i}}}.



Since {{{35/4}}} leaves a remainder of 3, this means the answer is {{{-i}}}.



So {{{i^35=-i}}}



So this means that 



{{{1/i^(35)=1/(-i)}}}





{{{1/(-i)}}}  Start with the given expression.




{{{(1/(-i))(i/i)}}} Multiply both numerator and denominator by "i" 



{{{(1*i)/((-i)(i))}}} Combine the fractions.



{{{(1*i)/(-i^2)}}} Multiply {{{i}}} and  {{{i}}} to get  {{{i^2}}} 



{{{(1*i)/(-(-1))}}} Replace {{{i^2}}} with {{{-1}}}. Remember {{{i^2=-1}}} 



{{{(i)/(1)}}} Multiply



{{{i}}} Simplify



So {{{i^(-35)=i}}}