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))}#