Trigonometric form of any complex number of the form a+iba+ib is given by r(cos(theta)+isin(theta)r(cos(θ)+isin(θ). If one is able to find rr and thetaθ then the job is done.
Let us see how we go about find rr and thetaθ let me teach you
Say our complex number is z=a+ibz=a+ib
We want to represent it as r(cos(theta)+isin(theta))r(cos(θ)+isin(θ)) let us distribute the rr first.
a+ib = rcos(theta)+isin(theta)a+ib=rcos(θ)+isin(θ)
Let us equate the real parts
a=rcos(theta)a=rcos(θ)
Squaring both sides.
a^2=r^2cos^2(theta)a2=r2cos2(θ)
Equating the imaginary parts we get
b=rsin(theta)b=rsin(θ)
Squaring both sides.
b^2=r^2sin^2(theta)b2=r2sin2(θ)
Adding a^2a2 and b^2b2
a^2+b^2 = r^2cos^2(theta)+r^2sin^2(theta)a2+b2=r2cos2(θ)+r2sin2(θ)
a^2+b^2 = r^2(cos^2(theta)+sin^2(theta))a2+b2=r2(cos2(θ)+sin2(θ)) 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"
Dividingb/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))}
