How do you express the complex number in trigonometric form: 5-5i?

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Answer 1 · 2018-04-06T12:09:43

#5sqrt(2)(cos(-pi/4)+isin(-pi/4))#

To convert from #z=x+iy# form to #z=r(costheta+isintheta)# form, you need to find #r# and #theta#.

#r=sqrt(x^2+y^2)# and #tantheta=y/x#

So, #r=sqrt(5^2+(-5)^2)=sqrt(50)=5sqrt(2)#

#tantheta=5/-5=>tantheta=-1=>theta=-pi/4#

#therefore 5-5i=5sqrt(2)(cos(-pi/4)+isin(-pi/4))#

Answer 2 · 2018-04-06T12:12:07

In trigonometric form: #7.07(cos 45-isin 45) #

#Z= a+ib=5 - 5i #. Modulus:#|Z|=sqrt(a^2+b^2) #

Modulus:#|Z|=sqrt(5^2+(-5)^2) #

#=sqrt 50 ~~ 7.07# Argument: #tan alpha=|b/a|= |5/-5|=1#

#:.alpha =tan^-1 (1) = 45^0 ; Z# lies on fourth

quadrant.#:. theta=360-alpha =315^0 or (-45)^0#

In trigonometric form expressed as

# Z=|Z|(cos theta+isin theta) #

# :. Z=7.07(cos 315+isin 315) # or

# Z=7.07(cos 45-isin 45) # [Ans]

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