Thread: Complex number and Logarithm help me

Solve for x and y real:
a) 7 - 2yi = 13x + 3i

b.) sin x + 1/2i = 1 + (cosy)i

solve for x:
a.) (log_4_(x+1))^2 - 3log_4_(x+1) - 4 = 0

_4_ is base
thanks!!!

Originally Posted by ^_^Engineer_Adam^_^

Solve for x and y real:
a) 7 - 2yi = 13x + 3i

rearrange:

13x+2y i = 7 - 3 i,

Now as x and y are real equate the real and imaginary terms on each side
or the equality to get:

13x=7, and 2y=-3,

or x=7/13, and y=-3/2.

RonL

Please double check the algebra.

Originally Posted by ^_^Engineer_Adam^_^

b.) sin x + 1/2i = 1 + (cosy)i

Like the last one:

sin(x) - i cos(y)=1 - (1/2) i,

so sin(x)=1, and cos(y)=1/2, which you
should be able to solve yourself

RonL

Originally Posted by ^_^Engineer_Adam^_^

solve for x:
a.) (log_4_(x+1))^2 - 3log_4_(x+1) - 4 = 0

_4_ is base

put y=log_4 (x+1), then the equation is:

y^2-3y-4=0

which has roots y=4, y=-1

So in the first case we have log_4 (x+1)=4, or:

x+1=4^4

x=4^4-1=256-1=255.

In the second case we have log_4 *x+1)=-1, or

x+1=4^-1,

x=3/4.

RonL

Again check the algebra yourself.