How do you write the complex number in standard form #3.75(cos((3pi)/4))+isin((3pi)/4))#?

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Answer 1 · 2016-08-31T14:33:43

#3.75(cos((3pi)/4)+isin((3pi)/4))=-2.652+i2.652#

As #cos((3pi)/4)=-1/sqrt2# and #sin((3pi)/4)=1/sqrt2#

#3.75(cos((3pi)/4)+isin((3pi)/4))#

= #3.75(-1/sqrt2+i×1/sqrt2)#

Now as #1/sqrt2=sqrt2/2# above is equal to

#3.75(-sqrt2/2+isqrt2/2)#

= #-(3.75×sqrt2)/2+i(3.75×sqrt2)/2#

= #-(3.75×1.4142)/2+i(3.75×1.4142)/2#

= #-2.652+i2.652#