What is the trigonometric form of # (5-i)*(3+i) #?

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Answer 1 · 2017-04-25T14:20:55

#2sqrt65 * ( cos theta + isin theta)#

where, #tan theta = 1/8 or theta = tan^-1(1/8)#

#(5-i)*(3+i)#

#= 5*3+5i-3i-i*i#

#=15+2i-sqrt(-1)*sqrt(-1)=15+2i-(-1)#

#=15+2i+1#

#=16+2i#

#=(16+2i)/(sqrt(16^2+2^2))*sqrt(16^2+2^2)#

#=[16+2i]/[sqrt(260)]*sqrt(260)#

#=2sqrt65[(16+2i)/(2sqrt65)]#

#=2sqrt65[8/sqrt65+i/sqrt65]#

#=2sqrt65[8/sqrt65+i*1/sqrt65]#

enter image source here

#=2sqrt65 * ( cos theta + isin theta)#

where, #tan theta = 1/8 or theta = tan^-1(1/8)#