What is the trigonometric form of # (2-25i) #?

1 Answer
Jim G.
Dec 17, 2016

#sqrt629(cos(1.49)-sin(1.49))#

Explanation:

The trigonometric form of a complex number #z=x+iy#

is #color(red)(bar(ul(|color(white)(2/2)color(black)(z=r(costheta+isintheta))color(white)(2/2)|)))#

#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(r=sqrt(x^2+y^2))color(white)(2/2)|)))#

#color(red)(bar(ul(|color(white)(2/2)color(black)(theta=tan^-1(y/x))color(white)(2/2)|)))#
where # -pi < theta <= pi#

Here #x=2" and " y=-25#

#rArrr=sqrt(2^2+(-25)^2)=sqrt629#

Now 2 - 25i is in the 4th quadrant, so we must ensure that #theta# is in the 4th quadrant.

#rArrtheta=tan^-1(-25/2)=-1.49larr" in 4th quadrant"#

#rArr2-25i=sqrt629(cos(-1.49)+isin(-1.49))#

which can also be expressed as.

#2-25i=sqrt629(cos(1.49)-isin(1.49))#

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