How is -1+i√3 written in polar form?

Please show steps

2 Answers
Daniel L.
Apr 26, 2018

See explanation.

Explanation:

First we need to calculate the module:

|z|=sqrt(a^2+b^2)=sqrt((-1)^2+sqrt(3)^2)=sqrt(1+3)=2|z|=√a2+b2=√(−1)2+√32=√1+3=2

Now we have:

z=|z|*(cosvarphi+isinvarphi)z=|z|⋅(cosφ+isinφ)

Here |z|=2|z|=2, so

cos varphi=(Re(z))/|z|=-1/2cosφ=Re(z)|z|=−12

The angle, for which cosvarphi=-1/2cosφ=−12 is varphi=120^oφ=120o, so now we can write the complex number in trigonometric form:

z=2*(cos120^o + isin120^o)z=2⋅(cos120o+isin120o)

Binayaka C.
Apr 26, 2018

In polar form expressed as 2(cos 2.0944+i sin 2.0944)2(cos2.0944+isin2.0944)

Explanation:

Let Z=a+i b ; Z=-1+ i sqrt 3 ; a= -1 ,b =sqrt 3 Z=a+ib;Z=−1+i√3;a=−1,b=√3 ;

Z=-1+ i sqrt 3 Z=−1+i√3 is in 2nd quadrant

Modulus |Z|=sqrt(a^2+b^2)=(sqrt((-1)^2+ (sqrt3)^2)) =2 |Z|=√a2+b2=(√(−1)2+(√3)2)=2

tan alpha =|b/a|= (sqrt 3/ -1) = sqrt 3 :. alpha =tan^-1(sqrt3)

or alpha~~ 1.047198; theta is on 2 nd quadrant

:. theta=pi- alpha= 2.094395 :. Argument : theta ~~2.0944

In polar form expressed as

|Z|*(cos theta+i sin theta) or 2(cos 2.0944+i sin 2.0944)[Ans]

Related questions
Impact of this question
1200 views around the world
You can reuse this answer
Creative Commons License