Question

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Answer 1

$\:\: \underline{\underline{\bf{\huge\mathcal{>formula~~ used<}}}}$

if ,

z=x=+iy

then ,the modulus-amplitude of the complex number z is ...

$\:\:\: \large\mathfrak{\underline{\underline{\large\mathcal{\bf{\boxed{\large\mathcal{Z=r(cos(theta)+i sin(theta))}}}}}}}$

==>Z=r(cos∅+i sin∅)

where...r=|z|=√(x²+y²)

and.......tan∅=y/X

$\:\: \underline{\underline{\bf{\large\mathfrak{~~Given~~}}}}$

$(Q1). \: \: \: \sqrt{3} + i \\ (Q2). \: \: \: \: 5 - 5i \\ (Q3). \: \: \: 4i \\ (Q4). \: \: \: ( - 3 + 3i)(1 - i) \\ (Q5). \: \: \: \frac{ \sqrt{3} - i}{1 - \sqrt{3}i }$

$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~Solution~~}}}}$

$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~(1)~~}}}}$

z1=√3+i

r1=|z1|=√(√3²+1²)=√4=2

tan∅=1/√3=>tan∅=tan30°=>∅=30°=π/6

therefore.mod-amp of z1 is..

$z1 = 2( \cos( \frac{\pi}{6} )+ i \sin( \frac{\pi}{6} ) )$

$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~(2)~~}}}}$

z2=5-5i=5+(-5)i

r2=|z2|=√(5²+(-5)²)=√50=5√2

tan∅=-5/5=-1=>tan∅=tan(-π/4)=>∅=-π/4

therefore.mod-amp of z2 is..

$z2 = 5 \sqrt{2} ( \cos( - \frac{\pi}{4} ) + i \sin( - \frac{\pi}{4} ) )$

$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~(3)~~}}}}$

z3=4i+0

r3=|z3|=√(√0²+4²)=√16=4

tan∅=0/4=>tan∅=tan0=>∅=0

therefore.mod-amp of z1 is..

Z3=4(cos0+isin0)

$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~(4)~~}}}}$

z4+=(-3+3i)(1-i)=-3+3i+3i+3=6i

r4=|z4|=√(√0²+6²)=√36=6

tan∅=0/6=>tan∅=tan0=>∅=0

therefore.mod-amp of z1 is..

Z4=6(cos0+isin0)

$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~(5)~~}}}}$

z5=(√3-i)/(1-√3i)=(√3-i)(1+√3i)/(1+3)

=(√3+3i-i+√3)/4=(2√3+2i)/4=(√3/2)+i(1/2)

r5=|z5|=√((√3/2)²+(1/2)²)=1

tan∅=(1/2)/(√3/2)=>tan∅=1/√3=>∅=π/6

therefore.mod-amp of z5 is..

Z5=(cos(π/6)+isin(π/6))

.

$:\: \underline{\underline{\bf{\large\mathfrak{~hope ~this ~help~you~~}}}}$

Answer 2

Answer:

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