<PRE><B>Question 725615</B>
1-i^34 divided by 3 square root -1 +5 (note that only -1 is under the square root and 3 is in front of it )
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i^34 = i^32*i^2 = 1*-1 = -1
----
(1-i^34)/(3i+5)
= 2/(5 + 3i)
----
Get the i out of the denominator
Multiply NUM and DEN by the conjugate of the DEN 5 - 3i
= 2(5-3i)/(5^2 + 9i^2)
= 2(5-3i)/16
= (5-3i)/8
or
5/8 - i*(3/8)
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