Trigonometric form of any complex number of the form #a+ib# is given by #r(cos(theta)+isin(theta)#. If one is able to find #r# and #theta# then the job is done.
Let us see how we go about find #r# and #theta# let me teach you
Say our complex number is #z=a+ib#
We want to represent it as #r(cos(theta)+isin(theta))# let us distribute the #r# first.
#a+ib = rcos(theta)+isin(theta)#
Let us equate the real parts
#a=rcos(theta)#
Squaring both sides.
#a^2=r^2cos^2(theta)#
Equating the imaginary parts we get
#b=rsin(theta)#
Squaring both sides.
#b^2=r^2sin^2(theta)#
Adding #a^2# and #b^2#
#a^2+b^2 = r^2cos^2(theta)+r^2sin^2(theta)#
#a^2+b^2 = r^2(cos^2(theta)+sin^2(theta))# #quadcolor(green)("factored out GCF " r^2#
#a^2+b^2 = r^2(1)##quadcolor(green) (quad cos^2(theta)+sin^2(theta)=1#
#a^2+b^2=r^2#
#=> r=sqrt(a^2+b^2)# #quad color(green) r# #color(green)" is a distance, so, it would be positive" #
Dividing#b/a# we get
#b/a = (rsin(theta))/(rcos(theta))#
#b/a = tan(theta)#
#=>tan^-1(b/a) = theta#
We get #theta = tan^-1(b/a)#
Now a question would arise is it needed to do so many steps. The answer to that it's not required .
We just need to know
#r = sqrt(a^2+b^2) quad theta=tan^-1 (b/a)#
Now let us get back to our problem.
#1+5i#
#r=sqrt(1^2+5^2)#
#r=sqrt(1+25)#
#r=sqrt(26)#
#theta = tan^-1(5/1)#
#theta = tan^-1(5)#
The trigonometric form
#1+5i = sqrt(26){cos(tan^-1(5))+isin(tan^-1(5))}#
