Question #36f01

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Answer 1 · 2017-11-04T12:40:37

In polar form expressed as #sqrt2(cos315+isin315)#

Let #Z=a+ib ; Z=1- i ; a=1 ,b =-1# ;

#Z=1+i# is in 4th qudrant

Modulus #|Z|=sqrt(a^2+b^2)=(sqrt(1^2+ (-1)^2)) =sqrt2 #

# tan theta =b/a= (-1)/1 or tan theta = -1 #

Argument # theta =tan^-1 (-1) = -45^0 or 315^0#.

In polar form expressed as #|Z|*(costheta+isin theta)#

#sqrt2(cos315+isin315)# [Ans]

Answer 2 · 2017-11-04T16:10:36

#sqrt2(cos(pi/4)-isin(pi/4))#

The polar form of a complex number

#z=a+ib#

is

#z=r(costheta+isintheta)#

where

#r=sqrt(a^2+y^2)#

#theta=arg(z) , -pi < theta <=pi#

we have
#z=1-i#

on an Argand diagram

enter image source here

#r=sqrt(1^2+1^2)=sqrt2#

#theta=tan^(-1)(_1/1)=-pi/4#

we have

#z=sqrt2(cos(-pi/4)+isin(-pi/4))#

cosine is an even function, and sine an odd one

#:.z=sqrt2(cos(pi/4)-isin(pi/4))#