Question

find x and y for e^z= 4i

Answer 1

what have you tried?

e^z = 4i

take natural log (ln) on both sides

Answer 2

x=1/2ln4 bt i m nt able to find y

Answer 3

z = x + iy

so the part without 'i' will be 'x'
and the part with 'i' will be y

Answer 4

z = ln(4i) = ln 4 + ln i = x + i y

and ln i = (pi/2) i

so , x = ln 2 or 2 ln 2
y = pi/2.

Answer 5

e(z)== 4i
i.e e^x(cosy i siny)= 4 i
i have to solve the given
equation by this method only

Answer 6

e(z)= 4i
i.e e^x(cosy+ i siny)= 4 i
i have to solve to the given
equation by this method only

Answer 7

here e^x.cosy =0 ___(1)
and e^x.siny=4____(2)
eqn (1)^2+(2)^2
e^2x= 16
x=ln4

Answer 8

ok then

e^x cos y + i e^x sin y = 0 +4i

comparing thee real and imaginary parts

e^x cos y = 0 ..... e^x sin y = 4

but e^x cannot be 0,
so
cos y = 0
so y = ...?

Answer 9

but m nt able to find solution of y

Answer 10

thn y will be (2n+1) pi/2

Answer 11

good work finding x! :)

Answer 12

x= ln 4 i knw tht bt
my teacher told me y =(2npi + pi/2) U( [2n+1)pi +pi/2}

Answer 13

find x and y for e^z= 2
here he told me value of y be {(2n+1)pi} U 2npi
m confused help me

Answer 14

cos y = 0 gives y=(2n+1) pi/2
but
sin y cannot be 0.

it will make imaginary part = 0.

Answer 15

bt in d given question imaginary prt is 0 ri8

Answer 16

not actually, 0+4i
0 is the real part,
4i is the imaginary part

Answer 17

y ops sry m mistake

Answer 18

here sin y = 1

Answer 19

hartnn thnks fr ur help :) :)