Question #173740

Let f(z)=z−1/z+1


. The questions that follow will all revolve around this complex-valued function.

1.

 The value of 25f(2+4i) is equal to a+ib where a,b∈R, and i^2=−1. What is the value of 10a+b?



1
Expert's answer
2021-03-23T05:33:06-0400
f(z)=z1z+1f(z)=z-\frac{1}{z}+1f(2+4i)=2+4i12+4i+1=f(2+4i)=2+4i-\frac{1}{2+4i}+1=

=2+4i24i(2+4i)(24i)+1==2+4i-\frac{2-4i}{(2+4i)(2-4i)}+1=

=2+4i24i4+16+1==2+4i-\frac{2-4i} {4+16} +1=

=2+4i24i20+1==2+4i-\frac{2-4i}{20}+1=

=2+4i0.1+0.2i+1==2+4i-0.1+0.2i+1=

=2.9+4.2i=2.9+4.2i

25f(2+4i)=25(2.9+4.2i)=72.5+105i25f(2+4i)=25(2.9+4.2i)=72.5+105i

a=72.5,b=105a=72.5, b=105


10a+b=1072.5+105=725+105=83010a+b=10\cdot{72.5}+105=725+105=830

Answer: 830


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