How do you divide #( 2i-9) / ( -7 i + 7 )# in trigonometric form?

1 Answer
Hari S.
Apr 27, 2018

#z=a+bi# = #r[z]e^(i*arctan(b/a))# by definition,
for #a,b,r[z] in RR#, #i = sqrt(-1)# , and #r[z] = sqrt(a^2+b^2)#
so for complex numbers #u=v+iw and z=a+bi#,
#(u/z)# =# (r[u])/(r[z]]# #e^(i(arctan(w/v)-arctan(b/a))#

In this case:
#u=2i-9# = #sqrt(2^2+9^2)*e^(i*arctan(-2/9))=9.2195e^(-.2187i)#
#z=7-7i = sqrt(2*(7^2))*e^(i*arctan(-1))=9.899e^(-.25ipi)#

So:
#((2i-9)/(-7i+7))# = #(9.2195/9.899)# #e^(i*(-.2187-.25*pi)#
= #0.9314*e^(-1.0041i)#

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