Solve [3+2i/4-3i]^2

$$[ \frac { 3 + 2 i } { 4 - 3 i } ] ^ { 2 }$$

Evaluate

$$-\frac{253}{625}+\frac{204}{625}i=-0.4048+0.3264i$$

Solution Steps

Multiply both numerator and denominator of \frac{3+2i}{4-3i} by the complex conjugate of the denominator, 4+3i.

$$\left(\frac{\left(3+2i\right)\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)}\right)^{2}$$

Do the multiplications in \frac{\left(3+2i\right)\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)}.

$$\left(\frac{6+17i}{25}\right)^{2}$$

Divide 6+17i by 25 to get \frac{6}{25}+\frac{17}{25}i.

$$\left(\frac{6}{25}+\frac{17}{25}i\right)^{2}$$

Calculate \frac{6}{25}+\frac{17}{25}i to the power of 2 and get -\frac{253}{625}+\frac{204}{625}i.

$$-\frac{253}{625}+\frac{204}{625}i$$

Real Part

$$-\frac{253}{625} = -0.4048$$

Solution Steps

Multiply both numerator and denominator of \frac{3+2i}{4-3i} by the complex conjugate of the denominator, 4+3i.

$$Re(\left(\frac{\left(3+2i\right)\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)}\right)^{2})$$

Do the multiplications in \frac{\left(3+2i\right)\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)}.

$$Re(\left(\frac{6+17i}{25}\right)^{2})$$

Divide 6+17i by 25 to get \frac{6}{25}+\frac{17}{25}i.

$$Re(\left(\frac{6}{25}+\frac{17}{25}i\right)^{2})$$

Calculate \frac{6}{25}+\frac{17}{25}i to the power of 2 and get -\frac{253}{625}+\frac{204}{625}i.

$$Re(-\frac{253}{625}+\frac{204}{625}i)$$

The real part of -\frac{253}{625}+\frac{204}{625}i is -\frac{253}{625}.

$$-\frac{253}{625}$$