How do you divide #( 2i-1) / ( -i -1 )# in trigonometric form?

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Answer 1 · 2018-07-02T11:53:26

#\sqrt{5/2}(\cos108.435^\circ-\sin 108.435^\circ)#

Given that

#\frac{2i-1}{-i-1}#

#=\frac{1-2i}{1+i}#

#=\frac{(1-2i)(1-i)}{(1+i)(1-i)}#

#=\frac{1-3i+2i^2}{1-i^2}#

#=\frac{1-3i+2(-1)}{1+1}#

#=\frac{-1-3i}{2}#

#=-1/2-3/2i#

Amplitude #=\sqrt{(-1/2)^2+(-3/2)^2}=\sqrt{5/2}#

Argument, #\theta=-(\pi-\tan^{-1}|\frac{-3/2}{-1/2}|)#

#\theta=-108.435^\circ#

#\therefore \frac{2i-1}{-i-1}#

#=\sqrt{5/2}(\cos(-108.435^\circ)+i\sin(-108.435^\circ))#

#=\sqrt{5/2}(\cos108.435^\circ-\sin 108.435^\circ)#