What is the trigonometric form of (1+5i) ?

1 Answer
Jan 20, 2016

1+5i = sqrt(26){cos(tan^-1(5))+isin(tan^-1(5))}

Explanation:

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"

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))}