Another Complex Number Question

**Stroodle** · July 30th 2009, 01:15 AM

If $z=x+yi$ determine the values of $x$ and $y$ such that $z=\sqrt{3+4i}$

**red_dog** · July 30th 2009, 01:40 AM

$3+4i=4+4i-1=4+4i+i^2=(2+i)^2$

Then $z_1=2+i, \ z_2=-2-i$

**Stroodle** · July 30th 2009, 01:48 AM

Cool. Thanks for your help.

**mr fantastic** · July 30th 2009, 03:06 AM

Originally Posted by Stroodle

If $z=x+yi$ determine the values of $x$ and $y$ such that $z=\sqrt{3+4i}$

Another approach:

$z^2 = 3 + 4i$

$\Rightarrow (x + iy)^2 = 3 + 4i$

$\Rightarrow (x^2 - y^2) + 2xy i = 3 + 4i$ .

Therefore:

$x^2 - y^2 = 3$ .... (1)

$2xy = 4 \Rightarrow xy = 2 \Rightarrow y = \frac{2}{x}$ .... (2)

Substitute (2) into (1): $x^2 - \frac{4}{x^2} = 3 \Rightarrow x^4 - 3x^2 - 4 = 0 \Rightarrow (x^2 - 4)(x^2 + 1) = 0$ .

But since x is real, $x = \pm 2$ .

x = 2: y = 1.
x = -2: y = -1.

**Stroodle** · July 30th 2009, 03:46 AM

Awesome. Thanks for that.
This method will be much easier for me to reproduce for similar problems.

Thread: Another Complex Number Question

LinkBack

Thread Tools

Display

Another Complex Number Question

Similar Math Help Forum Discussions

Complex number question

complex number question

can anyone help me complex number question?

Complex number question-

Search Tags