Solve 1/0.625-1.225i

$$\frac { 1 } { 0.625 - 1.225 i }$$

Evaluate

$$\frac{500}{1513}+\frac{980}{1513}i\approx 0.330469266+0.647719762i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 0.625+1.225i.

$$\frac{1\left(0.625+1.225i\right)}{\left(0.625-1.225i\right)\left(0.625+1.225i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{1\left(0.625+1.225i\right)}{0.625^{2}-1.225^{2}i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{1\left(0.625+1.225i\right)}{1.89125}$$

Multiply 1 and 0.625+1.225i to get 0.625+1.225i.

$$\frac{0.625+1.225i}{1.89125}$$

Divide 0.625+1.225i by 1.89125 to get \frac{500}{1513}+\frac{980}{1513}i.

$$\frac{500}{1513}+\frac{980}{1513}i$$

Real Part

$$\frac{500}{1513} = 0.3304692663582287$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{0.625-1.225i} by the complex conjugate of the denominator, 0.625+1.225i.

$$Re(\frac{1\left(0.625+1.225i\right)}{\left(0.625-1.225i\right)\left(0.625+1.225i\right)})$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{1\left(0.625+1.225i\right)}{0.625^{2}-1.225^{2}i^{2}})$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{1\left(0.625+1.225i\right)}{1.89125})$$

Multiply 1 and 0.625+1.225i to get 0.625+1.225i.

$$Re(\frac{0.625+1.225i}{1.89125})$$

Divide 0.625+1.225i by 1.89125 to get \frac{500}{1513}+\frac{980}{1513}i.

$$Re(\frac{500}{1513}+\frac{980}{1513}i)$$

The real part of \frac{500}{1513}+\frac{980}{1513}i is \frac{500}{1513}.

$$\frac{500}{1513}$$