the modulus of complex number......

[Ganesh Ujwal]

the modulus of complex number z = \(\displaystyle \dfrac{3+4i}{1-2i}\) is:
A) -π
B) - π/2
C) π/2
D) π
explain procedure also with ur answer....

 

[Deleted member 4993]

the modulus of complex number z = \(\displaystyle \dfrac{3+4i}{1-2i}\) is:
A) -π
B) - π/2
C) π/2
D) π
explain procedure also with ur answer....

Don't understand - what is "ur".

Again you have shown no work - so we are not showing any work!

refer to advices given in your previous post - http://www.freemathhelp.com/forum/threads/87921-how-to-solve-this-problem

and you have posted in: https://nz.answers.yahoo.com/question/index?qid=20140811231113AAhluob

 

[HallsofIvy]

the modulus of complex number z = \(\displaystyle \dfrac{3+4i}{1-2i}\) is:
A) -π
B) - π/2
C) π/2
D) π
explain procedure also with ur answer....

1) The modulus of the complex number a+ bi is \(\displaystyle \sqrt{a^2+ b^2}\)
2) To change a fraction to the form a+ bi, multiply both numerator and denominator by the complex conjugate of the denominator: if the denominator is u+ vi, multiply both numerator and denominator by u- vi.

However the modulus of this number is NOT any of the options you give! Are you sure the problem does not ask for the argument of this number? The "argument" of the complex number a+ bi is \(\displaystyle arctan(b/a)\) if a is not 0. If a is 0 and b is positive the argument is \(\displaystyle \pi/2\). If a is 0 and b is negative the argument is \(\displaystyle -\pi/2\). (The complex number 0 does not have an "argument".)

 

[Deleted member 4993]

1) The modulus of the complex number a+ bi is \(\displaystyle \sqrt{a^2+ b^2}\)
2) To change a fraction to the form a+ bi, multiply both numerator and denominator by the complex conjugate of the denominator: if the denominator is u+ vi, multiply both numerator and denominator by u- vi.

However the modulus of this number is NOT any of the options you give! Are you sure the problem does not ask for the argument of this number? The "argument" of the complex number a+ bi is \(\displaystyle arctan(b/a)\) if a is not 0. If a is 0 and b is positive the argument is \(\displaystyle \pi/2\). If a is 0 and b is negative the argument is \(\displaystyle -\pi/2\). (The complex number 0 does not have an "argument".)

HoI - don't waste time trying to teach this guy - he is too busy to even write the problem.

He got his answer - to this problem and the first problem - at the other site. He did not have to lift a finger maintaining his busy schedule.

 

[jonah2.0]

As I posted in reply to his 1st problem (It got deleted mysteriously):

Solved. Thank you for that thought-provoking, intellectually invigorating mental exercise.

My sarcastic post gets deleted but this lazy juvenile troller is allowed to keep on trolling. Go figure.

Why not just ban him permanently?

 

[John Marsh]

Answer is non of the options

the modulus of complex number z = \(\displaystyle \dfrac{3+4i}{1-2i}\) is:
A) -π
B) - π/2
C) π/2
D) π
explain procedure also with ur answer....

Given z=(3+4i)/(1-2i)
z=[(3+4i)(1+2i)]/[(1-2i)(1+2i)]
z=(-5+10i)/(5)
z=-5/5+10i/5
z=-1+2i

Modulus of the complex number a+bi=√(a^2+b^2)

Modulus of z=√(-1^2+2^2)
z=√(1+4)
z=√5
c

Answer is non of the options ​