document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1165360>Question 1165360</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If Z1,Z2 are part of C such that Im(z1) = sqrt(3), |z2| = 2 and z1*z2 = 4(cos( 5pi/6 ) + sin( 5pi/6)i ). Find the values of Z1 and Z2.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #853099 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2138)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=853099\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This problem requires using the properties of complex numbers in both rectangular and polar forms, along with the geometric interpretation of multiplication.\r<BR>\n" );
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document.write( "Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.\r<BR>\n" );
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document.write( "### 1. Analyze the Given Information\r<BR>\n" );
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document.write( "* **$z_1$ in Rectangular Form:** We are given $\operatorname{Im}(z_1) = \sqrt{3}$.<BR>\n" );
document.write( "    $$z_1 = x_1 + i\sqrt{3}$$<BR>\n" );
document.write( "* **$z_2$ in Polar Form:** We are given $|z_2| = 2$.<BR>\n" );
document.write( "    Let $z_2 = 2(\cos\theta_2 + i\sin\theta_2)$.<BR>\n" );
document.write( "* **Product ($z_p$):** $z_p = z_1 z_2 = 4\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right)$.\r<BR>\n" );
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document.write( "### 2. Convert the Product to Rectangular Form\r<BR>\n" );
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document.write( "First, find the exact value of the product $z_p$.<BR>\n" );
document.write( "$$\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$<BR>\n" );
document.write( "$$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$\r<BR>\n" );
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document.write( "$$z_p = 4\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)$$<BR>\n" );
document.write( "$$z_p = -2\sqrt{3} + 2i$$\r<BR>\n" );
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document.write( "### 3. Use the Modulus Property of the Product\r<BR>\n" );
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document.write( "The modulus of the product is the product of the moduli: $|z_1 z_2| = |z_1| |z_2|$.<BR>\n" );
document.write( "From the product's polar form, $|z_1 z_2| = 4$.<BR>\n" );
document.write( "We are given $|z_2| = 2$.<BR>\n" );
document.write( "$$4 = |z_1| \cdot 2$$<BR>\n" );
document.write( "$$|z_1| = 2$$\r<BR>\n" );
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document.write( "### 4. Determine $z_1$\r<BR>\n" );
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document.write( "We have $z_1 = x_1 + i\sqrt{3}$ and $|z_1| = 2$.<BR>\n" );
document.write( "The modulus is calculated as $|z_1| = \sqrt{x_1^2 + y_1^2}$:<BR>\n" );
document.write( "$$2 = \sqrt{x_1^2 + (\sqrt{3})^2}$$<BR>\n" );
document.write( "$$4 = x_1^2 + 3$$<BR>\n" );
document.write( "$$x_1^2 = 1$$<BR>\n" );
document.write( "$$x_1 = \pm 1$$\r<BR>\n" );
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document.write( "Therefore, there are two possible solutions for $z_1$:<BR>\n" );
document.write( "$$\mathbf{z_{1a} = 1 + i\sqrt{3}} \quad \text{or} \quad \mathbf{z_{1b} = -1 + i\sqrt{3}}$$\r<BR>\n" );
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document.write( "### 5. Determine $z_2$ (using $z_2 = \frac{z_p}{z_1}$)\r<BR>\n" );
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document.write( "We use the relationship $z_2 = \frac{z_p}{z_1}$, where $z_p = -2\sqrt{3} + 2i$.\r<BR>\n" );
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document.write( "#### Case A: Using $z_{1a} = 1 + i\sqrt{3}$\r<BR>\n" );
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document.write( "To divide, it's simplest to use the polar form for $z_{1a}$.<BR>\n" );
document.write( "$|z_{1a}| = 2$.<BR>\n" );
document.write( "$$\arg(z_{1a}) = \arctan\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$$<BR>\n" );
document.write( "$$z_{1a} = 2\left(\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right)$$\r<BR>\n" );
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document.write( "The argument of the quotient is the difference of the arguments: $\arg(z_2) = \arg(z_p) - \arg(z_{1a})$.<BR>\n" );
document.write( "From $z_p = 4\left(\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right)$, we have $\arg(z_p) = \frac{5\pi}{6}$.\r<BR>\n" );
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document.write( "$$\theta_{2a} = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi}{6} - \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$$\r<BR>\n" );
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document.write( "Since $|z_2| = 2$:<BR>\n" );
document.write( "$$z_{2a} = 2\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right) = 2(0 + i\cdot 1)$$<BR>\n" );
document.write( "$$\mathbf{z_{2a} = 2i}$$\r<BR>\n" );
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document.write( "#### Case B: Using $z_{1b} = -1 + i\sqrt{3}$\r<BR>\n" );
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document.write( "Polar form for $z_{1b}$:<BR>\n" );
document.write( "$|z_{1b}| = 2$.<BR>\n" );
document.write( "$$\arg(z_{1b}) = \pi - \arctan\left(\frac{\sqrt{3}}{1}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$<BR>\n" );
document.write( "$$z_{1b} = 2\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right)$$\r<BR>\n" );
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document.write( "$$\theta_{2b} = \arg(z_p) - \arg(z_{1b})$$<BR>\n" );
document.write( "$$\theta_{2b} = \frac{5\pi}{6} - \frac{2\pi}{3} = \frac{5\pi}{6} - \frac{4\pi}{6} = \frac{\pi}{6}$$\r<BR>\n" );
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document.write( "Since $|z_2| = 2$:<BR>\n" );
document.write( "$$z_{2b} = 2\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right)$$<BR>\n" );
document.write( "$$\mathbf{z_{2b} = \sqrt{3} + i}$$\r<BR>\n" );
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document.write( "### Final Solution\r<BR>\n" );
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document.write( "There are two pairs of complex numbers $(z_1, z_2)$ that satisfy the given conditions:\r<BR>\n" );
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document.write( "$$\text{Pair 1: } \mathbf{z_1 = 1 + i\sqrt{3}} \quad \text{and} \quad \mathbf{z_2 = 2i}$$<BR>\n" );
document.write( "$$\text{Pair 2: } \mathbf{z_1 = -1 + i\sqrt{3}} \quad \text{and} \quad \mathbf{z_2 = \sqrt{3} + i}$$ </SPAN>\n" );
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