What is the trigonometric form of 12+4i ?

2 Answers
Shwetank Mauria
May 19, 2016

12+4i=4sqrt10(cosalpha+isinalpha), where tanalpha=1/3

Explanation:

a+ib can be written in trigonometric form re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta),
where r=sqrt(a^2+b^2).

Hence 12+4i=sqrt(12^2+4^2)[cosalpha+isinalpha]

= sqrt160[cosalpha+isinalpha]

= 4sqrt10[cosalpha+isinalpha],

where cosalpha=12/(4sqrt10)=3/sqrt10

and sinalpha=4/(4sqrt10)=1/sqrt10

or tanalpha=(1/sqrt10)/(3/sqrt10)=1/3

Jim G.
May 19, 2016

4sqrt10(cos(0.322)+isin(0.322))

Explanation:

Given a complex number z = x + yi , this can be written in trig. form as.

z=x+yi=r(costheta+isintheta)

where r=sqrt(x^2+y^2)

and theta=tan^-1(y/x)

here x = 12 and y = 4

rArrr=sqrt(12^2+4^2)=sqrt160=4sqrt10

and theta=tan^-1(4/12)≈0.322" radians or"18.43^@

rArr12+4i=4sqrt10(cos(0.322)+isin(0.322))or

12+4i=4sqrt10(cos(18.43)^@+isin(18.43)^@)

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