Feb 14, 2016 · 9 - 19i >distribute the brackets using FOIL ( or any method you use ) hence (4 + i )(1 - 5i ) = 4 - 20 i + i - 5i^2 [note : i^2 = (sqrt-1)^2 = -1 ] so 4 - 20i + i + 5 = 9 - 19i

Oct 24, 2016 · For 2-i2, the polar form re^(itheta) is given by (2, -2)=r(cos theta, sin theta). Here, r = sqrt(2^2+(-2)^2)=2sqrt2, cos theta =2/r=1/sqrt 2>0 and sin theta =-2/r=-1/sqrt2<0.

Aug 23, 2016 · #rArr(4-3i)(2-i)=5-10i# Answer link. Related questions. How do I multiply complex numbers?

Dec 20, 2016 · In trigonometric form expressed as 5(cos330+isin330) Z=a+ib . Modulus: |Z|=sqrt (a^2+b^2); Argument:theta=tan^-1(b/a) Trigonometrical form : Z =|Z|(costheta+isintheta ...

Feb 14, 2016 · Multiply by 1, written as the complex conjugate of the bottom of the fraction divided by itself, in this case (1-2i)/(1-2i): (2+3i)/(1+2i)*(1-2i)/(1-2i)=(8-i)/5 ...

Dec 8, 2017 · A and C are two points in,an argand diagram representing the complex numbers -5+2i and 1+i respectively. Given that the AC the diagonal of the square ABCD, How to ...

Dec 9, 2015 · -1 Rewrite i^14 as (i^4)^3xxi^2. If i=sqrt(-1), then i^2=-1. From here (i^2)^2=(-1)^2, so i^4=1. If i^4=1, then we can say that: i^14=(1)^3xxi^2 =i^2 =-1